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3 years ago

Calculate the speed of a bike rider who accelerates (from rest) for 5 seconds down a hill at an acceleration of 8

Physics

1 answer:

olya-2409 [2.1K]3 years ago

6 0

Answer:

speed=abs(v)=40ms^-1

Explanation:

acceleration, a = (v-u)/t

since initial velocity u=0 (at rest) and a=8ms^-2,

8=v/5

hence after 5 seconds, v=40ms^-1

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A man does 500 j work pushing a car a distance of 2m how much force does he apply

olasank [31]

Answer: 250 N

Explanation:

Use equation for work

W=F*d

d=2m

W=500J

F=?

-----------------------

W=Fd

F=W/d

F=500J/2m

F=250N

6 0

3 years ago

Read 2 more answers

A 120.0 kg crate is placed on a 15.00°

Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0

3 years ago

Potassium-40 has a half-life of approximately 1.25 billion years. Approximately how many years will pass before a sample of pota

Maksim231197 [3]

Answer:

3.75 billion years

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 1.25 billion years

Number of half-lives (n) = 3

Time (t) =?

The time taken for the sample of potassium-40 to contains one-eighth the original amount of parent isotope can be obtained as:

n = t / t½

3 = t / 1.25

Cross multiply

t = 3 × 1.25

t = 3.75 billion years.

Therefore, it will take 3.75 billion years for the sample of potassium-40 to contains one-eighth the original amount of parent isotope

8 0

2 years ago

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2× $10^{-6}$ J

(b) KE = 2× $10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0

3 years ago

So I have to write a motion story for physics. For the assignments you need to have:

zheka24 [161]

Answer:

Explanation:

3 0

2 years ago

Read 2 more answers