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antoniya [11.8K]
2 years ago
11

HeLp aSAp!!! If the speed and distance of an object are given, and I need to find the time, I will...

Physics
2 answers:
Helen [10]2 years ago
5 0

Answer:

B - Divide Speed and Distance

Explanation:

Hope it helps!!

:D

gayaneshka [121]2 years ago
5 0
B. devide the speed and distance .
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A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
goldfiish [28.3K]
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
3 0
2 years ago
In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
What connects the upper motor neurons to lower motor neurons?
skad [1K]

Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.

Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

7 0
3 years ago
Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti
son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

b)

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃        (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0
3 years ago
You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr
Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

v_r = 27 - 10 = 17 m/s

relative deceleration of car

a_r = -7 m/s^2

now the distance before they stop with respect to each other is given by

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2 *(-7)*d

d = 20.6 m

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

v_f^2 - v_i^2 = 2 a s

0^2  - 27^2 = 2 * (-7)* s

s = 52.1 m

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

v_f - v_i = at

0 - 27 = (-7) * t

t = 3.86 s

now the distance covered by truck in same time

d = 3.86 * 10 = 38.6 m

now after the car will stop its distance from the truck is

D = 25 + 38.6 - 52.1 = 11.5 m

<em>so the distance between them is 11.5 m</em>

6 0
3 years ago
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