All categories

3 years ago

Physics {deceleration}

Physics

1 answer:

seropon [69]3 years ago

7 0

Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)

You might be interested in

Please help me with both of them

8_murik_8 [283]

Answer:

101011010101010100101000101010100010011010100010100000101041204105210241012021012021021012022221222122122345788981633333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333311111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111114565656565+4+652147

21212121512

546213171549895465621324547998995656565656565656565722426579898541321447985331321

Explanation:

4 0

3 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

3 years ago

A Universe Question!

vfiekz [6]

I'm pretty sure this is A

4 0

3 years ago

The search for black holes involves searching for The search for black holes involves searching for Group of answer choices sing

trasher [3.6K]

Answer:

Large spherical regions from which no light is detected

Explanation:

A black hole is an object that has an extremely high density such that it possesses very powerful gravitational force that prevents the escape of all objects including light from it, and consumes nearby objects.

Due to the power of the gravitational force of a black hole, at the center, objects are infinitesimally compressed resulting in the inapplicability of the concept of space and time and the location is known as a singularity

Therefore, the search for black holes involves searching for <em>large spherical regions from which no light is detected</em>.

4 0

2 years ago

During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. Max, the goalie, caught the ball 60 m away from

amid [387]

Answer:

3.0 seconds

Explanation:

We can solve the problem by considering the horizontal motion of the ball only. In fact, the ball moves by uniform motion (constant speed) along the horizontal direction, since there are no forces acting in this direction. The horizontal speed of the ball is given by:

$v_x = v_0 cos \theta = (25 m/s)(cos 37^{\circ})=19.97 m/s$

and it does not change during the motion.

We also know that the ball travels a horizontal distance of d = 60 m, so we can find the time it takes to cover the distance by using the equation:

$t=\frac{d}{v}=\frac{60 m}{19.97 s}=3.0 s$

8 0

3 years ago

Read 2 more answers