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KIM [24]
3 years ago
5

I will mark brainliest! :D SO please help me M8

Physics
2 answers:
Kobotan [32]3 years ago
7 0

Answer: 100 J

Explanation: 1/2 5 x 2^2 = 100

Hope this made any sense.

steposvetlana [31]3 years ago
5 0

A 50-kg object should have 100J kinetic energy if it is moving at a velocity of 2 m/s.

Explanation: Just use the formula! Remember to:

Plug in the weight/mass to find <em>m</em>

Plug in the speed/velocity to find <em>v</em>

<em />

So the equation changes from    1/2 x M x V^2

to...

1/2 x 50 x 2^2

^

this is equal to 100.

You might be interested in
vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. suppose that a storm is in progress wit
Agata [3.3K]

Before we answer this question, let us first understand what alternate hypothesis is.

The alternative hypothesis is the hypothesis which is used in the hypothesis testing and this is opposite to the null hypothesis. This is the test hypothesis which is usually taken to be that the observations are the result of a real effect in an experiment.

In this case since what we want to set up is the statistical test to see if the waves are dying down, then this means we are trying to determine if the wave height are decreasing, so lesser than 16.4 feet. Therefore:

The alternative hypothesis would state                 (ANSWER)

Ha: μ less than 16.4 feet and P-value area is on the left of the mean.

 

While the null hypothesis is the opposite and would state

H0: mu equals 16.4 feet 

4 0
3 years ago
A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro
ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let u_1, u_2 , v_f be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

m=2.50\times 10^{4}

Conserving momentum

mu_1+3mu_2=4mv_f

4m+6m=4mv_f

v_f=2.5 m/s

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J

Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J

Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ

4 0
3 years ago
Which of these is NOT an effect of humor?
IgorLugansk [536]
Feelings of jealousy and envy
7 0
3 years ago
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