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olga55 [171]
3 years ago
15

Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther

from the center? Explain your answer.​
Physics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

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Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

6 0
3 years ago
Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre
Doss [256]

Answer:

4 A

Explanation:

We are given that

R_1=R_2=R_3=4\Omega

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Using the formula

\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}

R=\frac{4}{3}\Omega

V=IR

Substitute the values

V=12\times \frac{4}{3}=16 V

I_1=\frac{V}{R_1}=\frac{16}{4}=4 A

I_1=I_2=I_3=4 A

Hence, current flows through any one of the resistors is 4 A.

7 0
3 years ago
How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?
DiKsa [7]
<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
4 0
3 years ago
Read 2 more answers
A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

6 0
3 years ago
The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive
agasfer [191]
<span>I believe it's insulation.</span>
5 0
3 years ago
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