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3 years ago

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Physics

1 answer:

Reptile [31]3 years ago

6 0

Answer: its C

Explanation:

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The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

3 0

3 years ago

What is the speed of an electron that has been accelerated from rest through a potential difference of 1020 V?

sashaice [31]

<span>a. KE in electron volts is 1020 eV.
b. KE in Joules is e(1020) = (1.6022E-19)(1020) = 1.634E-16
c. KE = (1/2)mv^2, so v = sqrt[2*KE/m] = 18.94E6 m/s

note: m is the mass of an electron = 9.109e-31 kg

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>

8 0

3 years ago

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

topjm [15]

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, m = mass

k = spring constant

Put the value into the formula

$\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}$

$\omega=4.74\ rad/s$

We need to calculate the period of oscillation,

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4.74}$

$T=1.33\ sec$

Hence, The period of oscillation is 1.33 sec.

4 0

3 years ago

Need help will mark you the Brainliest

Hatshy [7]

The last one is correct (D)

8 0

3 years ago

If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th

alex41 [277]

Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s

6 0

3 years ago