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Alborosie
3 years ago
5

A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p

oint? The acceleration of gravity is 9.81 m/s 2 . Answer in units of How long does the ball take to hit the ground after it reaches its highest point? Answer in units of s
Physics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s

It takes 2.84403 seconds to reach the highest point

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

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An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration
zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Distance  = 53m

Time  = 5.2s

Unknown:

Acceleration  = ?

Solution:

This is a linear motion and we use the right motion equation;

        S = ut  + \frac{1}{2}at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

 Insert the parameters and solve;

       53  = (0x 5.2) +  \frac{1}{2} x a x 5.2

       53  = 2.6a

         a = \frac{53}{2.6}  = 20.4m/s²

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2 years ago
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faltersainse [42]
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8 0
3 years ago
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A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni
Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

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That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

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3 years ago
Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed
Ivanshal [37]

Answer:

Part a)

f_w = f_g = 4.57 \times 10^{14} Hz

Part b)

\lambda_w = 492 nm

\lambda_g = 437.3 nm

Part c)

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v_g = 2.0 \times 10^8 m/s

Explanation:

Part a)

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f = \frac{v}{\lambda}

f = \frac{3 \times 10^8}{656 \times 10^{-9}}

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Part b)

As we know that the refractive index of water is given as

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\lambda_w = \frac{\lambda}{\mu_w}

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\lambda_g = \frac{\lambda}{\mu_g}

\lambda_g = \frac{656 nm}{3/2}

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Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

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v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

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