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2 years ago

When sunlight strikes the side of a building, what form of energy is it

Physics

2 answers:

ddd [48]2 years ago

6 0

Answer:

A thermal energy

Explanation:

Can I please have brainliest.

Katena32 [7]2 years ago

5 0

Answer:

I think its thermal energy

Explanation:

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A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?

Vikentia [17]

We know v=u +at
Here u = 0
a= 1.5 units
t= 5 s
So v= 5×1.5 m/s = 7.5 m/s

8 0

4 years ago

Read 2 more answers

A gun has a muzzle speed of 150 m/s. Find two angles of elevation that can be used to hit a target 800m away.

lorasvet [3.4K]

Answer:

Explanation:

u = 150 m/s

R = 800 m

The formula for the horizontal range is

$R=\frac{u^{2}Sin2\theta }{g}$

where, θ is angle of projection.

$800=\frac{150^{2}Sin2\theta }{9.8}$

Sin2θ = 0.348

2θ = 20.4°

θ = 10.2°

The range is same for the complementary angles.

So, the other angle is 90 - 10.2 = 79.8°

4 0

3 years ago

You are thinking about adding a part-time job to your business? You could pay the employees the $6.50 per hour. How much would y

SCORPION-xisa [38]

Answer:

1.) 15hrs per week = $97.50

2.) 20hrs per week = $130

Explanation:

4 0

3 years ago

B all figures represents molecules

Volgvan

C. Figure 11 and figure 1v are both compounds

5 0

3 years ago

An 18 gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm,1.02 mm, carries a constant current o

larisa86 [58]

Answer:

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

$J = \frac {I}{A}$

where I: electric current and A: cross-sectional area of the copper wire

The cross-sectional area of the copper wire can be calculated by:

$A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2}$

Substituting the calculated area in the equation (1) we have:

$J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}}$

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

$v = \frac {J}{n q}$

where n: the free-electron density, q: module of the charge of the electron

$v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)}$

$v = 1.50 \cdot 10^{-04} \frac {m}{s}$

So, the drift speed is 1.50x10⁻⁴ m/s.

Have a nice day!

4 0

4 years ago

Read 2 more answers