Ask question

Ask question

All categories

3 years ago

13

Solar energy is a form of

1 answer:

Alex Ar [27]3 years ago

4 0

Answer:

Solar energy is a form of Electromagnetic energy

Explanation:

Solar energy is constantly flowing away from the sun and throughout the solar system. Solar energy warms the Earth, causes wind and weather, and sustains plant and animal life. The energy, heat, and light from the sun flow away in the form of electromagnetic radiation (EMR

You might be interested in

What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

3 years ago

Relationship between voltage, resistance and current in a circuit

Lena [83]

Answer:

V = I×R

where -

V = potential difference across

I = current flowing in the circuit

R = Equivalent Resistance in the circuit

8 0

3 years ago

Is 5g greater than 20.43 mg

worty [1.4K]

yes that's true 6g is larger

4 0

3 years ago

Read 2 more answers

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago