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den301095 [7]
3 years ago
5

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration? A. 2 m/s2 B.15 m/s2 C.30 m/s2

D.50 m/s2
Physics
2 answers:
Ainat [17]3 years ago
6 0
A. 2 m/s^2

40-10=30
30÷15 =2
Black_prince [1.1K]3 years ago
5 0

Answer:

Acceleration, a=2\ m/s^2

Explanation:

It is given that,

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds.

It means,

Initial velocity of the car, u = 10 m/s

Final velocity of the car, v = 40 m/s

Time taken, t = 15 seconds

We have to find the acceleration of the car. Mathematically, acceleration of the car is given by taking difference in velocities divided by time taken i.e.

a=\dfrac{v-u}{t}

a=\dfrac{40\ m/s-10\ m/s}{15\ s}

a=2\ m/s^2

Hence, this is the required solution.

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Answer:

x = - 1.4

Explanation:

-5=10x+2-5x     (subtract 5x from both sides)

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-5-2=5x     (subtract 2 from both sides)

-7=5x      (simplify)

x=-7/5     (divide both sides by 5)

x=-1.4    (simplify)

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If an objects speed is constant and the direction is in a straight line what type of velocity does the object have
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An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla
dusya [7]

Answer:

C). U_f = \frac{U_0}{2}

Explanation:

As we know that capacitance of a given capacitor is

C = \frac{\epsilon_0 A}{d}

now we know that energy stored in the capacitor plates

U_0 = \frac{Q^2}{2C}

here if all the dimensions of the capacitor plate is doubled

then in that case

C' = \frac{\epsilon_0 (4A)}{2d}

here area becomes 4 times on doubling the radius and the distance between the plates also doubles

So new capacitance is now

C' = 2C

so capacitance is doubled

now the final energy stored between the plates of capacitor is given as

U_f = \frac{Q^2}{2C'}

so the final energy is

U_f = \frac{Q^2}{4C}

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4 0
3 years ago
The magnitude of a magnetic field a distance 2.0 µm from a wire is 36.0 × 10-4 T How much current is flowing through the
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The answer & explanation for this question is given in the attachment below.

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3 years ago
Dan is gliding on his skateboard at 2.00 m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at
miss Akunina [59]

Answer:

The velocity of Dan is 1.13 m/s.

Explanation:

Given that,

Initial velocity of skateboard and Dan = 2.00 m/s

Velocity of skateboard = 7.00 m/s

Mass of Dan m= 40.0 kg

Mass of skateboard M= 7.00 Kg

Suppose How fast is Dan going as his feet hit the ground?

We need to calculate the initial velocity of Dan

Using conservation of momentum

(m+M)v_{i}=mv_{f}+Mv_{f}

Put the value into the formula

(40.0+7.00)\times2.00=40.0\times v_{i}+7.00\times7.00

v_{i}=\dfrac{(40.0+7.00)\times2.00-7.00\times7.00}{40.0}

v_{i}=1.13\ m/s

Hence, The velocity of Dan is 1.13 m/s.

4 0
3 years ago
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