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3 years ago

You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

1 answer:

4 0

List out all the variables that you do know;

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41
v=√47.41
v=6.88549 ≈ 6.9 ms⁻¹

Hope I helped :)

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One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s

tiny-mole [99]

Answer:

Explanation:

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3 years ago

A nuclear fission power plant has an actual efficiency of 32%. If 0.18 MW of power are produced by the nuclear fission, how much

7nadin3 [17]

Answer:

P₀ = 5.76 x 10⁻² MW

Explanation:

given,

efficiency of the power plant = 32%

Power produced by the nuclear fission = 0.18 MW

the power plant output = ?

using formula of efficiency

$\eta = \dfrac{P_0}{P}$

where P is the power produced in the power plant

P₀ is the power output of the power plant

$\eta = \dfrac{P_0}{P}$

$0.32 = \dfrac{P_0}{0.18}$

P₀ = 0.18 x 0.32

P₀ = 0.0576 MW

P₀ = 5.76 x 10⁻² MW

Power plant output is equal to P₀ = 5.76 x 10⁻² MW

8 0

4 years ago

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it stri

Anni [7]

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight sThe x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.16227766016838, since the stone is thrown down (and to the right).(b) the initial speed m/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed m/sangle °

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4 years ago

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f

serg [7]

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$

The number of deuterons is $2.1875\times 10^{15}$

8 0

3 years ago

Ling lives 2 miles from school. It took him 15 minutes to bike from school to home. The first half of the distance he biked at a

Fittoniya [83]

Explanation:

As we know that

time = distance/speed

The time used for firs half of the trip was

(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes

The last half of the trip will took 10 minutes, 1/6 hour.

Speed = distance/ time

(1 mil) = (1/6h) = 6 mil/h

so the speed for last half of the trip was = 6mph

the average speed was

(2mil)(1/4 hour) = 8 mil/hour

So the ling's average speed was 8mph.

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4 years ago