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3 years ago

WILL GIVE BRAINLIEST!!!

Physics

1 answer:

vovikov84 [41]3 years ago

3 0

Answer:

boulder-sized rocks that come from the asteroid belt

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Can Hockey players hold the puck with their open hand?

devlian [24]

Answer:

Yes they can stop or bat the puck with their open hand

Explanation:

8 0

2 years ago

a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses

sweet-ann [11.9K]

Answer:

Explanation:

Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .

Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²

Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²

Applying law of conservation of angular momentum

I₁ ω₁ = I₂ ω₂

Putting the values ,

3 x .75 = .27 x ω₂

ω₂ = 8.33 rad / s

New angular speed = 8.33 rad /s .

8 0

2 years ago

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.

Galina-37 [17]

Answer:

Change in specific internal Energy $=250\ \rm Btu/lb$

Explanation:

Given:

Mass of the gas, m=0.4 lb
Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\$

Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\$

So the final volume of the gas is calculated.

Work in any isotropic is given by w

$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.

6 0

2 years ago

A 3 x 10^-6 C charge is 5m away from a -2x 10^-6 C charge A. Attractive because one is positive the other one is negative B. Det

Marizza181 [45]

Answer:

0.0021576N

Explanation:

F=(k)(q1q2/r^2)

F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)

F=0.0021576N

5 0

1 year ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago