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3 years ago

NEED THIS ANSWERED ASAP!!!! IN MIDDLE OF TEST RN HELP PLZ

2 answers:

4 0

The answer should be A, it does not exhibit projectile motion and follows a straight path down the ramp.

It's not in the air or anything so we definitely know it's not a projectile :D.

Marina86 [1]3 years ago

3 0

The correct answer for this question would be option A. When Barry is conducting an experiment and rolls a tennis ball down a ramp, the statement that best describes the motion of the tennis ball is that, i<span>t does not exhibit projectile motion and follows a straight path down the ramp. Hope this helps.</span>

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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be

ladessa [460]

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

6 0

2 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

In class I mentioned that we used to show an actual ballistic pendulum where a rifle is shot into a heavy block to see how the b

Len [333]

Answer:

0.66 m/s

Explanation:

This is an inelastic collision, for that reason the conservation of momentum law allows us to determine the final velocity of both, the bullet and block together after the collision:

$m_{bullet} v_{bullet0} +m_{block} v_{block} =(m_{bullet}+m_{block})v_{f} \\v_{f} =\frac{m_{bullet} v_{bullet0} }{m_{bullet}+m_{block}} \\v_{f} =\frac{0.0047kg*678m/s}{4.8kg+0.0047kg}\\v_{f} =0.66 m/s$

3 0

3 years ago

What is the mechanical energy of a 1-kg ball that is rolling on the ground at 2 m/s?

ruslelena [56]

Answer: The Answer is 2J

I just took the quiz

Explanation:

3 0

3 years ago