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3 years ago

What force is needed to accelerate a 1,000-kilogram car from a stop to 5 m/s/s?

1 answer:

3 0

That depends on how soon you want it to reach 5 m/s/s. Without friction, ANY force will accelerate the car, like a mosquito pushing on it, but a Space Shuttle booster will accelerate it at a greater rate.

You might be interested in

What is the momentum of a 3.5-kilogram object moving with a velocity of 6 m/s east?

Ghella [55]

Momentum formula - P=MV
P=MV
P= (3.5 kg)(6 m/s) - multiply
P= 21 kg x m/s to the east
Hope this helps

3 0

3 years ago

A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.52 radians with the

vovangra [49]

Answer:

a = 8.951 m/s²

Explanation:

given,

angle = 0.52 radians

μ_s = 0.84

μ_k = 0.48

acceleration = ?

using

F + f = m a

mg sin θ + μk mg cos θ = m a

a = g sin θ + μk g cos θ

a = 9.8 x sin 0.52 + 0.48 x 9.8 x cos 0.52

a = 4.869 + 4.082

a = 8.951 m/s²

the magnitude of acceleration is a = 8.951 m/s²

8 0

3 years ago

Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.

Snezhnost [94]

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

Therefore the train has a constant acceleration of about 0.23 meters per second.

5 0

3 years ago

if you crash your car how could you decrease the damage to you or the car using the concept of impulse

kotykmax [81]

Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

3 0

3 years ago

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

-10 + 12 sin 60 + Fₓ = 0

Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

12 cos 60 - 6 + F_y = 0

F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N

Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

$F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}$

F = 0.39N

We use trigonometry for the angle.

$tan \theta = \frac{F_y}{F_x}$

tan θ= 0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

θ = 180 + θ'

θ = 180 + 0

θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

10+ 6 + 6 + F = 0

F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

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

a = $\frac{24}{6}$

a = 4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

v = v₀ - a t

Final velocity when stopped is zero

t = $\frac{0-v_o}{a}$

t = 100/4

t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

2 years ago