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nikklg [1K]
4 years ago
13

The diagram shows a student throwing a baseball horizontally at 25 meters per second from a cliff 45 meters above the level grou

nd
Approximately how far from the base of the cliff does the ball hit the ground
Physics
1 answer:
Nimfa-mama [501]4 years ago
4 0

The horizontal distance covered by the ball is 75.8 m

Explanation:

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

First of all, we consider the vertical motion to find the time of flight of the ball. Using the suvat equation:

s=ut+\frac{1}{2}gt^2

where

s = 45 m is the vertical displacement of the ball

t is the time of flight

u = 0 is the initial vertical velocity of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s

Now we know that the ball moves horizontally with a constant velocity of

v_x = 25 m/s

So, the horizontal distance covered by the ball during its flight is

d=v_x t = (25)(3.03)=75.8 m

So the ball lands 75.8 m far from the base of the cliff.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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The rotating loop in an AC generator is a square l on each side. It is rotated at frequency f in a uniform field of B. The gener
spin [16.1K]

Answer:

Explanation:

Area of square loop = L²

Flux Φ = area x magnetic field

= L²B

Frequency = f

angular velocity ω = 2πf

a )

Let at time t = 0 , the magnetic field is making 90 degree with the face of the loop

flux through loop = L²B

After time t , coil will turn by angle ω t = 2πft

Flux through the loop = L²B cosω t

Φ (t) = L²B cosω t

= L²B cos2πft

b )

emf induced e

= - d/dt [Φ (t)]

= - d/dt [ L²B cosω t]

= L²B ω sinω t

= L²B 2πf sin2πft

c )

current = e / R

(L²B ω/ R ) sinω t

Power delivered

P(t) = VI ,

VOLT X CURRENT

= AB ω sinω t X ( AB ω/ R ) sinω t

= L⁴B² 4π²f²/R sin²2πft

e )

torque = MB sinω t

τ(t) = i(L²B ) sinω t

= (L²B ω/ R ) sinω t x (L²B ) sinω t

= (L²B  )²ω/ R sin²ω t

= (L²B  )² 2πf/ R sin²2πft

8 0
4 years ago
An object of mass 700700 kg is released from rest 20002000 m above the ground and allowed to fall under the influence of gravity
qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}

The equation of motion

x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})

x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})

when x(t)=2000

2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s

The time taken is 59.503987 seconds

8 0
3 years ago
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dsp73
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4 years ago
An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a
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The line charge E-field Ec = λ/(2πr*e0),
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