All categories

3 years ago

A race car starts from rest and accelerates down a track at a rate of 3.0 m/s2

Physics

1 answer:

7nadin3 [17]3 years ago

3 0

<h3>Answer:</h3>

30 m/s

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Velocity (u) = 0 m/s

Acceleration of the Car (a) = 3 m/s²

Time Interval (t) = 10 seconds

<u>Speed of the Car After 10 seconds:</u>

From the First equation of motion:

v = u + at

replacing the given values

v = 0 + (3)(10)

v = 30 m/s

Hence, the car is moving at a velocity of 30 m/s after 10 seconds

You might be interested in

An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len

worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

7 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

4 years ago

Find the focal length of a lens of power-2.0D . What type of lens is this ?

olga_2 [115]

Answer: f = - 0.50 m, negative (diverging) lens

Explanation: D-value (diopter ) means 1/f, unit is 1/m. Thus -2.0D

Means f = ( 1/ -2.0 ) m = -0.50 m

3 0

3 years ago

A runner starts from rest and achieves a maximum speed of 8.85 m/s.

ycow [4]

Answer:

The appropriate answer is "0.9152 seconds".

Explanation:

The given values are:

Maximum speed,

v = 8.85 m/s

Acceleration,

u = 9.67 m/s²

Now,

⇒ $v = u+at$

By putting the values, we get

⇒ $8.85=0+9.67t$

⇒ $t=\frac{8.85}{9.67}$

⇒ $=0.9152 \ second$

5 0

3 years ago

On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1

Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m = W/g

Where:

W : Weight (N)

g : acceleration due to gravity (m/s²)

Data

W₁ =1.8 kN : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁ m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}$

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}$

a₂= -2.17 m/s² : acceleration of the worker

5 0

4 years ago