What is the net force of a 5.0 N force and an 7.0 N force acting on an object for each of the following conditions?
(a) The forces act in opposite directions. ( )N
(b) The forces act in the same direction.( )N
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A force of 2.4 N is exerted on a 6.2 g rifle bullet. What is the bullet's acceleration?
( ) m/s2
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What is the force in newtons acting on a 4.6 kg package of nails that falls off a roof and is on its way to the ground?
( )N
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A small car with a mass of 800 kg travels northward at 28 m/s.
Does the car have more or less momentum than a truck with a mass of 12000 kg traveling eastward at 15 m/s?
more
less
How much more or less?
times
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Two ice skaters stand together as illustrated in Figure (a) below. They "push off" and travel directly away from each other, the boy with a velocity of v = 0.33 m/s to the left. If the boy weighs 728 N and the girl weighs 465 N, what is the girl's velocity after they push off? (Consider the ice to be frictionless. Enter the magnitude only.)
m/s
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An asteroid in an elliptical orbit about the sun travels at 1.5 106 m/s at perihelion (the point of closest approach) at a distance of 3.3 108 km from the sun. How fast is it traveling at aphelion (the most distant point), which is 9.2 108 km from the sun?
m/s
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To find the net force in each scenario, we need to add or subtract the forces acting on the object.
(a) When the forces act in opposite directions, the net force is the difference between the two forces. So, subtract the smaller force from the larger force to find the net force.
Net force = 7.0 N - 5.0 N = 2.0 N
(b) When the forces act in the same direction, the net force is the sum of the two forces. So, add the two forces together.
Net force = 5.0 N + 7.0 N = 12.0 N
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To find the bullet's acceleration, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = ma).
Given:
Force (F) = 2.4 N
Mass (m) = 6.2 g = 0.0062 kg
Rearrange the formula to solve for acceleration (a):
a = F/m
Plug in the values:
a = 2.4 N / 0.0062 kg
a ≈ 387.10 m/s^2 (approx.)
Therefore, the bullet's acceleration is approximately 387.10 m/s^2.
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To find the force acting on the package of nails, we can use the equation for weight, which states that weight is equal to mass multiplied by acceleration due to gravity (W = mg).
Given:
Mass (m) = 4.6 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Plug in the values:
W = (4.6 kg) * (9.8 m/s^2)
W ≈ 45.08 N (approx.)
Therefore, the force acting on the package of nails is approximately 45.08 N.
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To compare the momentum of the car and the truck, we can use the equation for momentum, which states that momentum is equal to mass multiplied by velocity (p = mv).
Given:
Car mass (m1) = 800 kg
Car velocity (v1) = 28 m/s
Truck mass (m2) = 12000 kg
Truck velocity (v2) = 15 m/s
Calculate the momentum of the car:
p1 = (800 kg) * (28 m/s)
p1 = 22400 kg m/s
Calculate the momentum of the truck:
p2 = (12000 kg) * (15 m/s)
p2 = 180000 kg m/s
Comparing the magnitudes of the momenta:
p1 < p2
Therefore, the car has less momentum than the truck.
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To find the girl's velocity after the push off, we need to apply the law of conservation of momentum, which states that the total momentum of a closed system before an event is equal to the total momentum after the event.
Given:
Boy's velocity (v1) = -0.33 m/s (to the left, negative sign indicates direction)
Boy's weight (W1) = 728 N
Girl's weight (W2) = 465 N
Calculate the total initial momentum:
p_initial = m1 * v1
= (W1/g) * (-0.33 m/s)
= -728 N / 9.8 m/s^2 * (-0.33 m/s)
≈ 25.437 kg m/s (approx.)
Since the system is closed (no external forces acting other than their weights during push off), the total momentum after the push off must also be equal to 25.437 kg m/s.
Since the boy pushes off to the left, the girl will move in the opposite direction and her velocity (v2) will have the same magnitude as the boy's velocity but in the opposite direction.
Therefore, the girl's velocity after the push off is 0.33 m/s to the right.
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To find the asteroid's speed at aphelion, we can use the law of conservation of angular momentum, which states that the product of an object's mass, velocity, and radius is constant in a circular or elliptical orbit.
Given:
Velocity at perihelion (v1) = 1.5 * 10^6 m/s
Radius at perihelion (r1) = 3.3 * 10^8 km = 3.3 * 10^11 m (convert km to m)
Radius at aphelion (r2) = 9.2 * 10^8 km = 9.2 * 10^11 m (convert km to m)
Using the conservation of angular momentum:
m1 * v1 * r1 = m2 * v2 * r2
Solving for v2 (velocity at aphelion):
v2 = (m1 * v1 * r1) / (m2 * r2)
Plug in the values:
v2 = (1.5 * 10^6 m/s * 3.3 * 10^11 m) / (m2 * 9.2 * 10^11 m)
Since the mass (m2) does not affect the result, it cancels out in the equation.
v2 ≈ 495.65 m/s (approx.)
Therefore, the asteroid's speed at aphelion is approximately 495.65 m/s.