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3 years ago

A car speeds up from rest to 40 m/s in 5 s. Find the car speed at 2 s and overall

Physics

2 answers:

alex41 [277]3 years ago

8 0

Answer:

I would have to say that i feel that after 2 seconds it would be 8 m/s and the overall distance would be 200 m but don't quote me on it

Explanation:

Nadya [2.5K]3 years ago

3 0

If a car goes from rest to 40 m/s in 5 seconds it means the acceleration is 8 m/s/s. After 2 seconds the speed is 16 m/s and the overall distance = 1/2 x 40 x 5 = 100m

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A ball is thrown upward at time t=0 from the ground with an initial velocity of 60 m/s (~ 134 mph). Assume that g = 10 m/s2.At w

Shtirlitz [24]

Answer:

6 second

Explanation:

initial velocity of ball, u = 60 m/s

g = 10 m/s^2

Let the ball takes time t to reach at the maximum height

We know that at maximum height, the velocity of ball is zero.

v = 0 m/s

Use first equation of motion

v = u + gt

0 = 60 - 10 x t

t = 6 second

Thus, the ball takes 6 second to reach to maximum height.

7 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

3 years ago

Will the sun ever stop shining? If so, how and when would we know((there are 3 questions

Alinara [238K]

Answer:

Explanation:stop it cause the sun don't shine that long unless if it's summer cause the summer time is very hot it may be 100 degrees or lower

6 0

3 years ago

A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to

Vitek1552 [10]

Answer:

W=561.41 J

Explanation:

Given that

m = 51 kg

μk = 0.12

θ = 36.9∘

Lets F is the force applied by man

Given that block is moving at constant speed it mans that acceleration is zero.

Horizontal force = F cos θ

Vertical force = F sinθ

Friction force Fr= μk N

N + F sinθ = m g

N = m g - F sinθ

Fr = μk (m g - F sinθ)

For equilibrium

F cos θ = μk (m g - F sinθ)

F ( cos θ +μk sinθ) = μk (m g

Now by putting the values

F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10

F= 70.2 N

We know that Work

W= F cos θ .d

W= 70.2 x cos 36.9∘ x 10

W=561.41 J

3 0

3 years ago

The 3 important steps to a perfect forehand in tennis are : perfect swing, type of racket, and correct finish

Alex17521 [72]

Answer:

False

Explanation:

The 3 important steps to a perfect forehand in tennis are Preparation,Backswing and Swing and contact.

3 0

2 years ago