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2 years ago

What is needed to give a large boulder a large acceleration?

Physics

2 answers:

hoa [83]2 years ago

7 0

You need a humongous force that you can apply to it.

jok3333 [9.3K]2 years ago

6 0

More force needs to be applied

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Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and

masha68 [24]

Answer:

F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

fr = μ N

We define a reference system parallel to the floor

block B ( lower)

Y axis

N - W₁-W₂ = 0

N = W₂ + W₂

N = (M + m) g

X axis

F -fr = M a

for block A (upper)

X axis

fr = m a (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

F = (M + m) a (3)

a = $\frac{F}{ M+m}$

the friction force has the formula

fr = μ N

fr = μ (M + m) g

let's calculate

fr = 0.34 (2.0 + 0.250) 9.8

fr = 7.7 N

we substitute in equation 2

fr = m a

a = fr / m

a = 7.7 / 0.250

a = 30.8 m / s²

we substitute in equation 3

F = (2.0 + 0.250) 30.8

F = 69.3 N

5 0

2 years ago

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

2 years ago

Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who

slega [8]

Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time

Thus, to get the average velocity for the first half of the swim

V(average) = displacement of first trip/time taken on the trip

V(average) = 50/20

V(average) = 2.5m/s

Average velocity for the second half of the swim will be calculated in like manner, thus,

V(average) = 50/55

V(average) = 0.91m/s

Average velocity for the round trip will then be

V(average) = 0/75, [50+25]

V(average) = 0m/s

3 0

2 years ago

1. A diffraction grating with 5.000 x 103 lines/cm is used to examine the sodium

Nuetrik [128]

Answer:

0.0002°, 0.1691°, 0.338°

Explanation:

Difference between the two line = 5.97 * 10-⁸m

d = 1 / N

N = 5.0 * 10³

d = 2.0 * 10⁴m

nL = Nsin¤

For first order

588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤

Sin¤ = 2.944*10-³

¤ = sin-¹ 0.002944

¤ = 0.1687°

First order ¤ =

Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)

Sin-¹ (0.002947) = 0.1689°

Angular separation = 0.1689 - 0.1687 = 0.0002°

Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)

Second order ¤ = 0.3378°

Angular difference = 0.3378° - 0.1687° = 0.1691°

Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°

Angular difference = 0.5067° - 0.1687° = 0.338°

7 1

2 years ago

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a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem

VikaD [51]

W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

6 0

2 years ago

Read 2 more answers