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bearhunter [10]
3 years ago
12

If a space ship traveling at 1000 miles per hour enters an area free of gravitational forces, its engine must run at some minimu

m level in order to maintain the ships velocity. Select one: True False
Physics
1 answer:
valentina_108 [34]3 years ago
4 0
That is false, if there are no forces acting upon it then its momentum will never die off there for unless a gravitational force acted upon it it will travel at 1000 MPH forever
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A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)
Tasya [4]

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

7 0
3 years ago
A bulldozer and a Mini Cooper are involved in a head-on collision. Which one experiences a greater force
Pepsi [2]

Answer:

The mini Cooper will experience the greater force

Explanation:

Generally, a bulldozer has a greater mass compared to a Mini Cooper hence when both of these vehicles interact in an head on collision the Mini Cooper will experience a greater force because the bulldozer has a greater momentum

5 0
3 years ago
Can someone help with thsi? i will give brainliest
kogti [31]

Answer:

40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.

Explanation:

4 0
3 years ago
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
Please help I'll mark brainliest!!!!
madam [21]

This question is based on the fundamental assumption of  vector direction.

A vector is  a physical quantity which has  magnitude as well direction  for its complete specification.

The magnitude of a physical quantity is simply a  numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector.  For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion  is negative and vertically upward is positive. In case of a falling object the motion is  vertically downward. So the velocity of that object is negative .

So last   option is  partially  correct  as  the vector can be negative depending on our choice of co-ordinate system.





7 0
3 years ago
Read 2 more answers
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