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3 years ago

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What is the instantaneous velocity at time=60 seconds

1 answer:

Mumz [18]3 years ago

3 0

Answer:i don't knowExplanation:

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If you add together all of forces exerted on an object and get a non zero value this is called the _____ force on the object

butalik [34]

Your answer would be the; <u>NET</u> force on the object. Refer to Newton's Laws of Forces and Motion.

Hope that helps!!!!!!!!!!!!!!!! : )

8 0

3 years ago

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Force is measured in units called

balu736 [363]

Answer:

Newtons

Explanation:

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4 years ago

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A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

tangare [24]

Answer:

Explanation:

Given

speed of Electron $u=2\times 10^7\ m/s$

final speed of Electron $v=4\times 10^7\ m/s$

distance traveled $d=1.2\ cm$

using equation of motion

$v^2-u^2=2as$

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

$(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}$

$a=5\times 10^{16}\ m/s^2$

acceleration is given by $a=\frac{qE}{m}$

where q=charge of electron

m=mass of electron

E=electric Field strength

$5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}$

$E=248.3\ kN/C$

5 0

3 years ago

In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

Nata [24]

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

6 0

3 years ago