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Xelga [282]
2 years ago
7

A force of 20 N acts on a rocket for 350 s, causing the rocket's velocity to increase. Calculate the impulse of the force and by

how much does the rocket's momentum increase?
Physics
1 answer:
Alex73 [517]2 years ago
7 0
<span>The formula for impulse is force times time, so impulse would be 20N x 350s = 7000 N x s. Since this impulse of 7000 N x s, the momentum would also be the same since the velocity increased after this impulse.</span>
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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

5 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
3 0
2 years ago
A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H
ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

M=\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the size of the image

h_o = 2.0 m is the height of the real object (the man)

q=50 mm =0.050 m is the distance of the image from the lens

p = 5.0 m is the distance of the object (the man) from the lens

Solving the formula for h_i, we find

h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm

And the negative sign means the image is inverted.

6 0
2 years ago
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