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Svetach [21]
3 years ago
13

Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

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2 years ago
honor physics The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force. a. Wha
Ksivusya [100]

Answer:

(a) 4000 kgm/s.

(b) 80 N

Explanation:

(a) Change in momentum: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is Ns or kgm/s

Mathematically, Change in momentum is expressed as

ΔM = mΔv ..................................... Equation 1

Where ΔM = change in momentum, m = mass of the object, Δv = change in velocity = v₂ - v₁

Given: m = 200 kg, Δv = v₂ - v₁ = 25-5 = 20 m/s.

Substituting into equation 1

ΔM = 200(20)

ΔM = 4000 kgm/s.

Hence the change in momentum = 4000 kgm/s

(b)

Force: This can be defined as the ratio of the change in momentum of a body to the time required for the change.

F = ΔM/t.............................. Equation 2

Where F = force, ΔM = change in momentum, t = time.

Given: ΔM  = 4000 kgm/s, t = 50 second.

Substituting into equation 2

F = 4000/50

F = 80 N.

Hence the force  = 80 N

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3 years ago
two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m
alexdok [17]
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges: 

<span>F α (1/2)Q *(1/2)q
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3 years ago
Read 2 more answers
A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,
Lorico [155]

Answer:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

             y = y₀ + v_{oy} t - ½ g t²

             0 =y₀ + 0 -1/2 gt²

             t = \sqrt{ \frac{2y_o}{g} }

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

          v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = \sqrt{v_x^2 + v_y^2}

          v = \sqrt{12^2 + 24.5^2}

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

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2 years ago
A 35 grams bullet travels with a velocity of magnitude 126 km/h. What is the bullet's linear momentum?
LekaFEV [45]

Answer:

IDK

Explanation:

idk

3 0
3 years ago
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