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Vika [28.1K]
3 years ago
7

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

ulate a) The acceleration of the box (assuming it is constant while sliding)
Physics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v=  0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

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An airplane travels 3,260 kilometers in 4 hours. What is the airplane’s average speed?
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Explain: a camel walks easily on sand surface than a man inspite of the fact that a camel is much heavier than a man
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4 0
3 years ago
Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w
Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor =  3.53 V

4 0
3 years ago
In the diagram, q1 = +6.39*10^_9 C and
ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

q_1=6.39\times 10^{-9} \ C

q_2=3.22\times 10^{-9} \ C

AP=(0.150+0.250)

      =0.40 \ m

BP=0.25 \ m

Now,

At point P, the electric potential will be:

⇒ V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}

By putting values, we get

⇒     =9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]

⇒     =259.695 \ Volt

4 0
3 years ago
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