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olga nikolaevna [1]
3 years ago
6

A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses the buildi

ng on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect air resistance.
With what speed was the rock thrown?
How high did the rock travel in the reference frame of the thrower?
Physics
1 answer:
Vlada [557]3 years ago
4 0

Explanation:

Given that,

Distance, d = -56.3 m

It strikes the ground 4.00 seconds after being thrown.

Using second equation of motion to find the speed was the rock thrown. So,

d=ut+\dfrac{1}{2}at^2

Here, a = -g

d=ut-\dfrac{1}{2}gt^2\\\\-56.3=u(4)-4.9(4)^2\\\\-56.3=4u-78.4\\\\u=5.52\ m/s

Let it will cover a distance of s meters. So,

s=\dfrac{v^2-u^2}{-2g}\\\\s=\dfrac{0^2-(5.52)^2}{-2\times 9.8}\\\\s=1.55\ m

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Answer:The electric field is zero and the potential is positive.

Explanation:

Two identical positive charges are separated by a certain distance and midway between charges two identical positive charges are placed near each other.

So the Electric field at midway is zero because the electric field due to both charges add up to give zero electric field.(because they point in opposite direction)

Potential is scalar quantity and charges are positive so they add up to give potential.

7 0
3 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

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3 0
3 years ago
You are given an unknown liquid and asked to determine its identity. You can measure only one physical property. You decide to h
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A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc
Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

<u>Given the following data;</u>

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

<em>Angle = 41.7° Northeast.</em>

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

<em>Vr = 7.08m/s</em>

6 0
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Answer:

a

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