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Vanyuwa [196]
3 years ago
10

How can the current through the material be changed? thx :)

Physics
1 answer:
Rina8888 [55]3 years ago
6 0
It depends on what the material is
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The amount of heat needed to increase the temperature of a 1-kilogram substance by 1°C is known as the
den301095 [7]

The amount of heat needed to increase the temperature of a 1-kilogram substance by 1°C is known as the specific heat of the substance.

the formula for specific heat of a substance is given as

c = Q/(m ΔT)

where Q = Heat required to change the temperature by 1°C

m = mass of the substance

ΔT = change in temperature.

the units of specific heat is given as Joules/(kilogram °C)


5 0
3 years ago
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2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

8 0
3 years ago
Physical Science AIS - AE 20-21-Kotal / Unit 2: Reactions and Radioactivity / The Environment
astraxan [27]

Answer:19

Explanation:

6 0
3 years ago
A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o
marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, m=5.0 kg

Force acting towards east, F=27 N

Frictional force acting towards west, f=17 N

Let the acceleration be a m/s².

Now, net force acting on the box towards east is given as:

F_{net}=F-f=27-17=10\textrm{ N}

From Newton's second law of motion,

F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2

Therefore, the magnitude of the acceleration of the box is 2 m/s².

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3 years ago
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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

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3 years ago
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