All categories

2 years ago

A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?

Physics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

You might be interested in

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car

saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

$F = ma$

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

$-10000 =2000\times a$

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

$a = -\dfrac{10000}{2000}$

$a=-5\ m/s^2$

For the distance,

Using third equation of motion

$v^2-u^2=2as$

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

$0-30^2=2\times(-5)\times s$

$s = 90\ m$

Hence, The car traveled the distance before stopping is 90 m.

6 0

3 years ago

Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto

kodGreya [7K]

Answer:

R = (- 3.72î + 8.29j)

Magnitude of R = 9.09 m

Explanation:

Let î and j represent unit vectors along the x and y axis respectively.

Vector A --> magnitude 8.78 m, direction 37.0° from the +x-axis

Let the x and y components of this vector be Aₓ and Aᵧ

A = (Aₓî + Aᵧj) m

The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

Bₓ = (8.26 cos 135°) = - 5.84 m

Bᵧ = (8.26 sin 135°) = 5.84 m

B = (-5.84î + 5.84j) m

Vector C has magnitude 5.65 m and direction 210° from the +x-axis

C = (Cₓî + Cᵧj) m

Cₓ = (5.65 cos 210°) = - 4.89 m

Cᵧ = (5.65 sin 210°) = - 2.83 m

C = (- 4.89î - 2.83j) m

The resultant force is a vector sum of all the forces. Let the resultant force be R

R = (Rₓî + Rᵧj) m

R = A + B + C = (7.01î + 5.28j) + (-5.84î + 5.84j) + (- 4.89î - 2.83j)

Summing the î and j components seperately,

R = (- 3.72î + 8.29j) m

To get its magnitude,

Magnitude of R = √(Rₓ² + Rᵧ²) = √((-3.72)² + (8.29)²) = 9.09 m

8 0

4 years ago

3 characteristics of chemical change

gtnhenbr [62]

Answer:

A physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.

3 0

3 years ago

Read 2 more answers

The scientist who suggested that energy can be created under certain conditions was

tester [92]

Einstein hope this helped

7 0

4 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLIEST!!!

Elan Coil [88]

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

$\frac{1}{2}kx^2 = mgh$

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

$g=9.8 m/s^2$ is the acceleration due to gravity