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3 years ago

Answer the question fast please

Physics

1 answer:

AnnZ [28]3 years ago

6 0

Answer:

Explanation:

$d = m \div v \\ m = 0.8 \: v = 80\% \: also \: 0.8 \\ d = 0.8 \div 0.8 \\ d = 1$

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3 years ago

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

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3 years ago

A wave travels at 330m/s^-1. the wavelength is found to be 2.4m.

sveta [45]

Answer: frequency 137.5

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Explanation:

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2 years ago

A golf ball reaches a height of 150 m before it stops rising and starts to fall to the ground. What is the golf balls speed (rou

artcher [175]

Answer:

v = 54 m/s

Explanation:

Given,

The maximum height of the flight of golf ball, h = 150 m

The velocity at height h, u = 0

The velocity of the golf ball right before it hits the ground, v = ?

Using the III equations of motion

Substituting the given values in the above equation,

v² = 0 + 2 x 9.8 x 150 m

= 2940

v = 54 m/s

Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s

4 0

3 years ago

A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t

Shalnov [3]

Answer:

$C=\frac{ab}{k(b-a)}$

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

$C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\$

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

$dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]$

Hence, the capacitance is

$C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}$

but R1=a and R2=b

$C=\frac{ab}{k(b-a)}$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers

Answer the question fast please​

Answer the question fast please