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4vir4ik [10]
4 years ago
6

A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

me the net braking force acting on it has magnitude 8.01×10−16 N and is directed opposite to its initial velocity. Over what distance must the force be applied? Watch your negative signs in this problem.
Physics
1 answer:
Tpy6a [65]4 years ago
8 0

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

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From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0
3 years ago
How much physical activity do adults need
Irina18 [472]
For most healthy adults, the Department of Health and Human Services recommends these exercise guidelines: Aerobic activity. Get at least 150 minutes a week of moderate aerobic activity or 75 minutes a week of vigorous aerobic activity. You also can do a combination of moderate and vigorous activity.
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Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
gizmo_the_mogwai [7]

Answer:

Mike can travel 80 Km in 4 hours

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3 years ago
A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of
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Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

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3 0
2 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

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

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

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3 years ago
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