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Evgen [1.6K]
3 years ago
11

What is the energy equivalent of an object with a mass of 1.83 kg? 1.65 × 1017 J 2.92 × 1017 J 3.10 × 1017 J 5.49 × 1017 J

Physics
2 answers:
Charra [1.4K]3 years ago
6 0

Answer:

1.65 x 1017j

Explanation:

Kryger [21]3 years ago
5 0

Answer:

Explanation:

We can use the Einsten's relation which states the equivalence between mass and energy:

where

E is the energy

m is the rest mass of the object

c is the speed of light

By using this formula and m=2.5 kg, we can find the equivalent energy of the object:

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lianna [129]

The first choice on the list is the correct one.

7 0
3 years ago
The frequency of the second harmonic of a certain musical instrument is 100 Hz. What is the fundamental frequency of the instrum
ruslelena [56]
The harmonic frequency of a musical instrument is the minimum frequency at which a string that is fixed at both ends in the instrument may vibrate. The harmonic frequency is known as the first harmonic. Each subsequent harmonic has a frequency equal to:
n*f, where n is the number of the harmonic and f is the harmonic frequency. Therefore, the harmonic frequency may be calculated using:
f = 100 / 2
f = 50 Hz
4 0
3 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
3 years ago
The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i
Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0
3 years ago
What is the force required to accelerate a baseball with a mass of 0.145 kilograms and an acceleration of 35.00 meters/second sq
Paladinen [302]

Answer:5.075N

Explanation:

Mass=0.145kg

Acceleration=35m/s^2

Force=mass x acceleration

Force=0.145 x 35

Force=5.075N

3 0
3 years ago
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