If an astronaut weighs 900 n on earth, what does he weigh on the planet venus?

How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The norm

DIA [1.3K]

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

8 0

3 years ago

Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?

alexandr402 [8]

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

7 0

3 years ago

Physics question, please show work for brainliest :)

Wittaler [7]

The answer is D. 4200J

Because 10 x 420 = 4200

8 0

3 years ago

Read 2 more answers

Why are experiments often performed in laboratories?

kap26 [50]

In my opinion,I think the answer is b.You can control variables more easily by doing different things for different purposes.

4 0

3 years ago

Read 2 more answers

Projectile <br> SHOW WORK<br> WILL MARK BRANLIEST <br> (Draw Picture and Label)

m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

a)

The motion of the projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

b)

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height: