All categories

2 years ago

A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?

Physics

2 answers:

stepladder [879]2 years ago

7 0

Answer:

a= 0.256 ft/s^2

Explanation:

Solution:

Since the car has started from rest the initial velocity will be zero

i.e. Vi=0

It gets a speed of 1.6 fts^2,

i.e. Vf= 1.6 fts^2

The distance is 5 yards

i.e d= 5 yards

According to the equation of motion

2as= VF^2- Vi^2

Thus a= (VF^2- Vi^2)/2s

a= (1.6^2 - 0)/ 2*5

a= 0.256 ft/s^2

Nina [5.8K]2 years ago

5 0

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

You might be interested in

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

2 years ago

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

Black_prince [1.1K]

Answer:

a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of the cylindrical shell

A = π r_b² - π r_a²

A = π (r_b² - r_a²)

we substitute

R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

V = i R

we subsist the expression of resistance

V = I ρ₀ L /π (r_b² - R_a²)

ρ₀ = V / I π (r_b² - R_a²) / L

6 0

2 years ago

Why is pluto not a planet?

mihalych1998 [28]

Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.

Hope this helps!

3 0

3 years ago

Read 2 more answers

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$