Friday

September 4, 2015
Total # Posts: 612

**Physics**

How much torque is applied to a bicycle wheel whose radius is 0.40 m in order to accelerate the 15 kg bike and 78 kg rider at a rate of 3.4 m/s^2? I'm totally lost on this one. Which equation do I need to use?
*January 8, 2008*

**Physics**

I still cannot solve this problem: Consider a spaceship located on the Earth-Moon center line (i.e. a line that intersects the centers of both bodies) such that, at that point, the tugs on the spaceship from each celestial body exactly cancel, leaving the craft literally ...
*December 17, 2007*

**Physics**

I've tried this several times, and I keep coming up with 43765.73 km, which is not correct. ???
*December 17, 2007*

**Physics**

Consider a spaceship located on the Earth-Moon center line (i.e. a line that intersects the centers of both bodies) such that, at that point, the tugs on the spaceship from each celestial body exactly cancel, leaving the craft literally weightless. Take the distance between ...
*December 17, 2007*

**Physics**

Oh you're right, I forgot that step. So 5.07E-12 N is the net force with both balls.
*December 17, 2007*

**Physics**

(6.67E-11)(2.16)(0.0088) / (0.5068)^2 = 4.94E-12 N Is that correct?
*December 17, 2007*

**Physics**

Two 2.160 kg crystal balls are 0.870 m apart in the vertical direction. Calculate the magnitude of the gravitational force they exert on a 8.80 g marble located 0.260 m horizontally out from the center of a line that intersects the centers of the balls. (For your own knowledge...
*December 17, 2007*

**Physics**

Ohhh I see now! Sorry, I just wasn't understanding before. Thanks. :)
*December 16, 2007*

**Physics**

Right. But w/o knowing the actual mass of the earth and moon, only the ratio, how can I do this?
*December 16, 2007*

**Physics**

Suppose that the ratio of the Moon's mass to the Earth's mass is given by 1.200E-2 and that the ratio of the Moon's radius to the Earth's radius is given by 2.700E-1. Calculate the ratio of an astronaut's Moon-weight to Earth-weight. Now it seems like this ...
*December 16, 2007*

**Physics**

Actually, that just made me even more confused. I still don't know what I'm supposed to use as the distance from the spaceship to the earth. Thanks, tho.
*December 16, 2007*

**Physics**

Ok I understand that. But what is the distance from craft to earth?
*December 16, 2007*

**Physics**

Oh wait...do I need to use F = GM1M2/d^2?
*December 16, 2007*

**Physics**

Consider a spaceship located on the Earth-Moon center line (i.e. a line that intersects the centers of both bodies) such that, at that point, the tugs on the spaceship from each celestial body exactly cancel, leaving the craft literally weightless. Take the distance between ...
*December 16, 2007*

**Physics**

Great I got it, thanks for all your help.
*December 15, 2007*

**Physics**

My last problem for the day: A) The centers of two 11.00 kg spheres are separated by 0.09 m. What is their gravitational attraction? I got 9.96 x 10^-7 N. B) What is the ratio of this attraction to the weight of one of the spheres (at the surface of the Earth)? I'm sure ...
*December 15, 2007*

**Physics**

Oh ok I see. Well, my online homework site would only accept it as postive, but it has been known to make errors before.
*December 15, 2007*

**Physics**

I tried both of the ways and got the same answer. However, the answer I got using bobpursley's equation got me -5.11E-8, when it was really positive. Was there an error in that equation, or did I just mess up with the math somewhere along the way?
*December 15, 2007*

**Physics**

Three very small spheres are located along a straight line in space far away from everything else. The first one (with a mass of 2.43 kg) is at a point between the other two, 10.60 cm to the right of the second one (with a mass of 5.17 kg), and 20.70 cm to the left of the ...
*December 15, 2007*

**Physics**

The Wall of Death in an amusement park is comprised of a vertical cylinder that can spin around the vertical axis. Riders stand against the wall of the spinning cylinder and the floor falls away leaving the riders held up by friction. The radius of the cylinder is 3.8 m and ...
*December 15, 2007*

**Physics**

This is the info for the problem: A block of mass M1 = 160.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and frictionless pulley. The coefficient of static friction between the block and the plane is ms = 0.52, and the angle...
*December 12, 2007*

**Physics**

Ahhh ok that was what was wrong. Thanks a ton. :)
*December 9, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. I still get 0.315 m as the answer, but it's ...
*December 9, 2007*

**Physics**

Ok I got it, thanks. The last part is this: Suppose the coefficient of kinetic friction between the block and the inclined plane is muk = 0.36. If M2 = 217.1 kg, what is the magnitude of the acceleration of M1? Which equation am I using for this?
*December 9, 2007*

**Physics**

That's right. Now it asks for the minimum value of M2 for which the system will remain at rest. Do I still use mgSinTheta-mg*mu*CosTheta?
*December 9, 2007*

**Physics**

A block of mass M1 = 160.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and frictionless pulley. The coefficient of static friction between the block and the plane is ms = 0.52, and the angle between the plane and the ...
*December 9, 2007*

**math**

49% is 0.49 as a decimal, and 80% is 0.80 as a decimal. 0.49 x 0.80 = 0.392, or 39.2%
*December 8, 2007*

**Physics**

Using 21.6 as the angle for both cos and sin, I come up with 33.91, which is the incorrect answer. Am I still doing something wrong?
*December 8, 2007*

**Physics**

Why is it sin and cos of 23.5? I thought the angle was 21.6...
*December 8, 2007*

**Physics**

Do you come up with an answer of 33.91? Because that's what I get.
*December 8, 2007*

**Physics**

Ok I understand that part. Now it asks for the minimum magnitude of F that will start the block moving up the plane, and the magnitude of F that is required to move the block up the plane at constant velocity. Is it a similar process for both?
*December 8, 2007*

**Physics**

A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping...
*December 8, 2007*

**Physics**

I've entered this into the computer, but it says it's not right. Hmmmm...
*December 6, 2007*

**Physics**

I keep getting 0.315 m, but that's not right. What am I doing wrong?
*December 6, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. So I get that (1/2)MV^2 = MgX*sin 15 + Mg*muk*...
*December 6, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. So I get that (1/2)MV^2 = MgX*sin 15 + Mg*muk*...
*December 5, 2007*

**Physics**

What is V, in this case?
*December 5, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. I still don't quite understand this one. How...
*December 5, 2007*

**Physics**

Ahh ok, I didn't realize the masses cancel out. But of course you're right. That really was easy. :)
*December 5, 2007*

**Physics**

A hockey puck is sliding across a frozen pond with an initial speed of 6.5 m/s. It comes to rest after sliding a distance of 7.6 m. What is the coefficient of kinetic friction between the puck and the ice? I know this one is not that difficult. However, I can't seem to get...
*December 5, 2007*

**Physics**

Yeppp n/m it is correct. Thanks a lot. :)
*December 5, 2007*

**Physics**

I came up with 5079 N as my answer to a. Is that correct?
*December 5, 2007*

**Physics**

A 850.0-kg car travelling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). (a) Find the size of the horizontal force which the car applies on the road while stopping. (b) Find the stopping distance of that same car when...
*December 5, 2007*

**Physics**

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? I'...
*December 5, 2007*

**Physics**

Ohh ok great I understand. Thanks a lot. :)
*December 4, 2007*

**Physics**

A 5.70 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.4 N at an angle theta = 27.5° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.10. What is the speed of the block 6.1 s after it ...
*December 4, 2007*

**Physics**

So wait...do I use cos 15 to find the acceleration?
*December 4, 2007*

**Physics**

I come up with 0.233 m as my answer, but that's not right. What am I doing wrong?
*December 4, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. Can someone please show me the steps to solving ...
*December 4, 2007*

**Physics**

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. I'm never sure which equation to use for ...
*December 3, 2007*

**Physics**

See, I thought that was right, too. But I plug that into my computer and it says it's not right. So I'm not sure what I'm doing wrong...
*December 3, 2007*

**Physics**

A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping...
*December 3, 2007*

**Physics**

A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping...
*December 3, 2007*

**Physics**

A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping...
*December 2, 2007*

**Physics**

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? Ok so...
*December 2, 2007*

**Physics**

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? Ok so...
*December 1, 2007*

**Physics**

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? Ok so...
*November 30, 2007*

**Physics**

How do I find the mass?
*November 30, 2007*

**Physics**

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? I ...
*November 30, 2007*

**Physics**

For lift I got 1320, and for deltamass I got 304.76. Is this right? If so, do I have to add those to the netforce?
*November 28, 2007*

**Physics**

A research balloon of total mass 200 kg is descending vertically with a downward acceleration of 3.2 m/s^2. How much ballast must be thrown from the car to give the balloon an upward acceleration equal to 2.8 m/s^2, presuming that the upward lift of the balloon does not change...
*November 28, 2007*

**Physics**

N/m I finally got it. Thanks for all your help. :)
*November 27, 2007*

**Physics**

I come up with 0.96 m/s^2. Is this correct?
*November 27, 2007*

**Physics**

m is 68 kg?
*November 27, 2007*

**Physics**

A man stands on a scale in an elevator that is accelerating upward. The scale reads 731.6 N. When he picks up a 35.0 kg box, the scale reads 1108.2 N. The man weighs 68 kg. What is the acceleration of the elevator? What do I need to do in order to solve this?
*November 27, 2007*

**Physics**

A car that weighs 14900.0 N is initially moving at a speed of 57.0 km/hr when the brakes are applied and the car is brought to a stop in 4.8 s. Find the magnitude of the force that stops the car, assuming it is constant. I found a using Vf=vi+at...a = 11.88 m/s^2. Then I found...
*November 26, 2007*

**Physics**

Now it asks for the normal force exerted on the mass by the floor. Do I need to use the acceleration for that?
*November 26, 2007*

**Physics**

A block of mass 7.5 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 31.0 N at an angle 26.3° above the horizontal. What is the magnitude of the acceleration of the block? I have nooo clue what to do for this one. Can someone please show ...
*November 26, 2007*

**Chemistry**

C2H6 = 14 valence electrons There are 4 valence electrons in a C, so in 2 C there is 8. H has only one. 8 + (1 x 6) = 14 v.e.
*November 14, 2007*

**Chemistry**

17? I come up with 19, but you might want to double check that.
*November 14, 2007*

**Physics repost**

I come up with 13.6 N as my net force. Is this correct?
*November 14, 2007*

**Physics repost**

A 2 kg otter starts from rest at the top of a muddy incline 85 cm long ans slides down to the bottom in 0.5 s. What net external force acts on the otter along the incline? I know which equation I need to use. But how can I use the cm and s to get the acceleration?
*November 14, 2007*

**Physics**

A 2 kg otter starts from rest at the top of a muddy incline 85 cm long ans slides down to the bottom in 0.5 s. What net external force acts on the otter along the incline? I know which equation I need to use. But how can I use the cm and s to get the acceleration?
*November 14, 2007*

**Chemistry**

My worksheet gives me a list of chemical symbols, and each one makes a phrase. I have to figure out which each one says. The one I can't figure out has these symbols: Co, Mo, Tc, O, U, N, Ta, Re, Y Any guesses as to what the phrase could be?
*November 13, 2007*

**Physics**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 12, 2007*

**Physics**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 12, 2007*

**Physics**

Great, thanks. That helps a lot. But now I have to do a graph of velocity and the x/y directions. My velocity in the x-direction is a staight line, and the vel. of the y continually decreases. They ask me to explain why each line has the shape that it does in terms of slope ...
*November 11, 2007*

**Physics**

Ahh ok I see. There isn't anything I should add to them? I just want to make sure they are as complete as possible.
*November 11, 2007*

**Physics**

We're studying projectile motion in my Physics class right now, and last week we went outside and hit tennis balls with a bat. Now we must make graphs showing the position of the ball in both the x and y directions. I must also explain why each line has the shape that it ...
*November 11, 2007*

**Physics**

16.73 is the final?? Because again, it's not right. Hmm...
*November 9, 2007*

**Physics**

The soccer goal is 23.05 m in front of a soccer player. She kicks the ball giving it a speed of 17.97 m/s at an angle of 25.83 degrees from the horizontal. If the goalie is standing exactly in front of the net, find the speed of the ball just as it reaches the goalie. ...
*November 9, 2007*

**Physics - check my work**

I finally got the right answer. And I was using the right equation for Vfy, just not using the right numbers.
*November 9, 2007*

**Physics - check my work**

Hmm yeah I figured Vfy was wrong. But I'm still a little confused on which I equation I DO need to use.
*November 9, 2007*

**Physics - check my work**

A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground? Please check my work. I think it’s wrong, but I don’t know where I’m ...
*November 9, 2007*

**Math**

You might have to correct me if I'm wrong, but I think... In order to calculate overtime pay, you must take your amount earned per hour and multiply it by 1.5. In your case, Eric's overtime rate would be $13.86/hour ($9.24 x 1.5 = $13.86). For double time, take the ...
*November 9, 2007*

**Algebra**

There really is no formula for this. All you must do is add 8 to both sides. What you do to one side, you must do to the other. w-8 = 9 w-8(+8) = 9(+8) The eights cancel out on this side. Now you are simply left with w = 17. Make sense?
*November 9, 2007*

**Physics repost**

I keep getting the same answer as you. This is what I did to get my original answer: V^2 = sqrt[Vx^2 + Vy^2] where... Vx = 41 m/s * cos 38 Vy at impact can be calculated using Vy^2 = (41 sin 38)^2 + 2 g H I got 55.56 as my Vy. I have a feeling this is where I got messed up...
*November 9, 2007*

**Physics repost**

Would 59.5 be the final answer?? B/c that is not right, either...
*November 9, 2007*

**Physics repost**

A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground? I keep getting 64.28 m/s as my answer, but it's not right. Could someone tell me ...
*November 9, 2007*

**Physics**

I keep getting 64.28 m/s as my answer, but it's not right. What am I doing wrong?
*November 8, 2007*

**Physics**

V is 41 m/s, right?
*November 8, 2007*

**Physics**

A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground? Which equations do I need for this?
*November 8, 2007*

**Physics repost**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 8, 2007*

**Physics**

Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court? What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???
*November 8, 2007*

**Physics**

Ok so I have the initial speed and the max height. Now it asks "At what initial speed must the ball be hit so that it lands directly on the opponent's back line?" Do I use a process similar to that of the first question?
*November 8, 2007*

**Physics**

Now it asks for the max. height. above the court reached by the ball. Do I use the equation Vf^2 = vi^2 + 2(g)(delta y) ?
*November 8, 2007*

**Physics**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 8, 2007*

**Physics**

This is probably really obvious, but I'm still not undertanding why you have to subtract one.
*November 7, 2007*

**Physics**

How did you come up with 39.96 Vh^2?
*November 7, 2007*

**Physics**

The launching speed of a certain projectile is 6.4 times the speed it has at its maximum height. Calculate the elevation angle at launching. I have noo idea how to start this one. Please help!
*November 7, 2007*