A force of 70.0 N is exerted at an angle of 30.0° below the horizontal on a block of mass 8.00 kg that is resting on a table. What is the magnitude of the normal force acting on the block?

normal force= mg+70sin30

113.48

Well, let me calculate that for you. But first, let me ask you a question - why did the scarecrow win an award? Because he was outstanding in his field!

To find the magnitude of the normal force, we can use a little bit of trigonometry. The force exerted at an angle of 30.0° below the horizontal can be split into two components - one acting horizontally and one acting vertically. The vertical component of the force will be responsible for the normal force.

The vertical component can be found using the formula: F_vert = F * sin(angle). Plugging in the values, we get:

F_vert = 70.0 N * sin(30.0°)
F_vert = 35.0 N

The normal force acting on the block is equal in magnitude but opposite in direction to the vertical component of the force. Therefore, the magnitude of the normal force is 35.0 N.

I hope my witty answer didn't make you fall off your seat! If you need any further assistance, feel free to ask!

To find the magnitude of the normal force acting on the block, we need to consider the forces acting on the block.

1. The force of gravity or weight acting downwards. The weight can be calculated using the formula: W = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
W = 8.00 kg * 9.8 m/s^2 = 78.4 N.

2. The force exerted at an angle of 30.0° below the horizontal. This force can be divided into two components: the horizontal component and the vertical component. The vertical component will contribute to the normal force acting on the block.

The vertical component can be calculated using the formula: F_vert = F * sin(θ), where F is the magnitude of the force and θ is the angle.

F_vert = 70.0 N * sin(30.0°)
F_vert = 70.0 N * 0.5
F_vert = 35.0 N

3. The normal force acting on the block is equal in magnitude and opposite in direction to the vertical component of the force. Therefore, the normal force is 35.0 N (downwards).

So, the magnitude of the normal force acting on the block is 35.0 N.

To find the magnitude of the normal force acting on the block, we need to understand the forces involved in this situation.

The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface and prevents the object from falling through the surface.

In this case, the weight of the block is directed downward due to gravity, and the force of 70.0 N is acting at an angle of 30.0° below the horizontal.

To find the normal force, we first need to resolve the force of 70.0 N into its horizontal and vertical components.

The vertical component can be found by multiplying the force by the sine of the angle:
Vertical component = 70.0 N * sin(30.0°)
Vertical component = 35.0 N

The horizontal component can be found by multiplying the force by the cosine of the angle:
Horizontal component = 70.0 N * cos(30.0°)
Horizontal component = 60.6 N

Now that we have the horizontal and vertical components, we can analyze the forces in the vertical direction.

The weight of the block acts vertically downward and is given by:
Weight = mass * acceleration due to gravity
Weight = 8.00 kg * 9.8 m/s^2
Weight = 78.4 N

The vertical component of the applied force (35.0 N) and the weight force (78.4 N) must balance for the block to remain at rest. Therefore, the magnitude of the normal force would be the sum of these two forces:
Normal force = Vertical component of applied force + Weight
Normal force = 35.0 N + 78.4 N
Normal force = 113.4 N

Therefore, the magnitude of the normal force acting on the block is 113.4 N.