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2 years ago

PLEASE help with this question!

Physics

1 answer:

Ainat [17]2 years ago

6 0

Answer:

the answer might the number 2

Explanation:

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Please help. Having a hard time figuring out

Goryan [66]

Yeah that’s is correct

5 0

3 years ago

When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

$\lambda = \frac{c}{f}$

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

$f = 2Hz$

$c = 3*10^8km/s$

Replacing we have,

$\lambda = \frac{c}{f}$

$\lambda = \frac{3*10^8km/s}{2Hz}$

$\lambda = 1.5*10^8m$

<em>Therefore the wavelength of this wave is $1.5*10^{8}m$ </em>

8 0

3 years ago

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

3 years ago

UCaShoTLLOuuj9D4vqFtTpSQ<br><br> search

Bond [772]

Answer:

ok i searched it up in google and it showed links XDDDD

Explanation:

4 0

3 years ago