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Goshia [24]
2 years ago
14

What is the acceleration of a rocket if 200 Newtons are applied to its 5.5 kg​

Physics
1 answer:
snow_tiger [21]2 years ago
7 0
Given
Force=200N
Mass=5.5kg

F=ma
a=F/m=200/5.5=38.18metre per second
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Pls help i will give u brianliest!
Fofino [41]

Answer:

Explanation:

neautrons and protons

3 0
3 years ago
An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
erik [133]

Answer:

Bounce 1 ,  pass 3,   emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle.  So it is  Bounce1, pass3, emb2.  

6 0
3 years ago
Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?
Molodets [167]
Taking notes,because you can read back on what you learned
4 0
3 years ago
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w
Scilla [17]

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

I = mR^2

now we will have

KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}

KE = mv^2

now by work energy theorem we can say

W = KE_f - KE_i

842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

3 0
2 years ago
Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on
Tamiku [17]

Explanation:

In the given question, the two metal spheres were hanged with the nylon thread.

When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.

The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.

During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.

It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.

4 0
3 years ago
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