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3 years ago

A boat moves initially at 10 m/s, then accelerates at 2 m/s^2 for 10 s. What is the velocity of the boat after 10 s?

Physics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

Explanation:

acceleration is the change in velocity of a body with respect to time

a = v-u/t

v is the final velocity

u is the initial velocity = 10m/s

t is the time taken = 10s

a is the acceleration = 2m/s²

substitute the values into the formula

2 = v-10/10

cross multiply

20 = v-10

v = 20+10

v = 30m/s

Hence the velocity of the body after 10s is 30m/s

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A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the li

suter [353]

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is $n_1=1.33$

Refractive index of the air that is $n_2=1$ (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law $n_1sini=n_2sinr$

$1.33sin32.7=1\times sinr$

$sinr=0.7185$

r =45.93°

So the light leave the water at an angle of 45.93°

3 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

frozen [14]

this is false, they can lift up to 10,000 pounds at a time

4 0

3 years ago

Read 2 more answers

A car has a uniform velocity of 108km/h. How far does it travel in 30 seconds?

natita [175]

Velocity=108km/h

Convert to m/s

$\\ \tt\longmapsto 108\times \dfrac{5}{18}$

$\\ \tt\longmapsto 30m/s$

Time=30s.

$\\ \tt\longmapsto Distance=Velocity\times Time$

$\\ \tt\longmapsto Distance=30(30)$

$\\ \tt\longmapsto Distance=900m$

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3 years ago

You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The

AnnZ [28]

Answer:

0.75

Explanation:

Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.

Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).

So, μ = F/N

= 300 N/400 N

= 3/4

= 0.75

So, the coefficient of static friction μ = 0.75

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2 years ago

The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei

Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+6450} \right )^2$

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+33700} \right )^2$

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0

3 years ago