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Advocard [28]
2 years ago
5

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s. What is the veloc

ity of car A after the collision? Would greatly appreciate the actual answer!
Physics
1 answer:
aalyn [17]2 years ago
6 0

Answer:

Hello There. ☆~\《--_^■^_--》\~☆ The final velocity of the car A is -1.053 m/s. For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Hope It Helps!~ ♡

ItsNobody~ ☆

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In the introduction of this activity, we mentioned the temperature of your home on hot and cold days. Your body is kept warm in
Alecsey [184]

Answer:

a - The walls of the house are warmer during the summer. Therefore, they radiate more infrared light that can serve to warm our body.

Explanation:

The answer is Option(a). According to studies, the reason to feel warmer inside the house is the absorption of infrared radiations emitted by the walls. Also more the temperature of the body, the more the heat radiated by it. During summer the walls of the house are at higher temperature for which they emit more infrared radiations which our body absorbs and make us feel more warmer.

3 0
2 years ago
Read 2 more answers
a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum
7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 ×10^{-2} m

height (h) =8.8 cm = 8.8 ×10^{-2} m

total charge (q) =6.1×10^{-7} C

To solve:

volume charge density (ρ)

We know;

<u> ρ =q ÷ v</u>

volume of cylinder = π ×r  × r × h

volume of cylinder =3.14 × 2.1 × 2.1 ×10^{-4} × 8.8 ×10^{-2}

volume of cylinder (v) = 122.23 ×10^{-6}

<u> ρ =q ÷ v</u>

ρ = 6.1×10^{-7} ÷   122.23 ×10^{-6}

<u>ρ = 0.04 × </u>10^{-1}<u> C</u>

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

3 0
2 years ago
A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3
garik1379 [7]

Answer:

A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

 Part (a) Express the wavelength of the light in the oil,  λ₀, in terms of  λ  and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of  λ o

(c) Express tmin in terms of  λ  and no.

(d) Solve for the numerical value of tmin in nm.

Explanation:

n₀ = 1.47

refraction of water = 1.3

refraction of air = 1

wavelength λ = 775 nm

(a) wavelength of light in water ⇒  λ₀ = λ / n₀

(b) minimum thickness of the film that will result in destructive interference

t(min) = λ₀ / 2

(c) the express t(min)

t =  λ /2n₀

(d) the thickness is

t = 775 / 2(1.47)

= 263.61 nm

4 0
2 years ago
Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder
Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0
2 years ago
A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie
Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0
2 years ago
Read 2 more answers
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