What is the momentum of a 5100 kg car has a velocity of 12 m/s ?

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

A player kicks a football from ground level with an initial velocity of 27.0 30.0 above the horizontal Find the ball's maximum h

stepladder [879]

Answer:

Given

initial velocity (u) =27.030

Force of gravidity g) =9.8

Rtc maximum height Hmix =?

$sln \\ \ hmix = u {}^{2} \div 2g \\ 27.030 \div 2 \times 9 \\ \\ hmix = 38.025m$

5 0

2 years ago

Exercise can help prevent?

Sedaia [141]

health conditions and diseases

8 0

3 years ago

Read 2 more answers

QUESTION 3

Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, $\mu_s$ = 0.5

Wood on wood kinetic friction, $\mu_k$ = 0.3

Solution;

The force of friction, $F_f$ , acting when the crate is moving is given as

follows;

$F_f$ = m × g × $\mu_k$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

$F_f$ = 100 × 9.81 × 0.3 = 294.3

The force of friction, $F_f$ = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

$a = \dfrac{F}{m}$

Therefore;

$a = \dfrac{293.7 \ N}{100 \ kg} = 2.937$

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0

2 years ago

A vehicle has an oil leak that is causing the entire oil pan to be wet, but inspection reveals no exact source after cleaning of

attashe74 [19]

Answer:

Technician A is correct

Explanation:

The best approach to solve the problem is that of technician A. using a fluorescent die is the easiest and most efficient way to trace leaks with unknown sources. The fluorescent die will simply illuminate the path to the leaking spot in the engine of the car, without any need for much speculations. This makes this method a sure approach.

However, Technician B's approach still has a lot of assumptions factored into the methodology, and would not work properly. It will still require the painstaking attempts trying to make guesses where the oil leak is coming from, which will lead to wastage of time and energy.

This makes Technician A have the right approach to solving the problem

7 0

3 years ago