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3 years ago

A graph shows how the temperature of a substance changes as energy is added steadily over time.

Physics

2 answers:

OlgaM077 [116]3 years ago

8 0

Answer:

A flat, horizontal line

Explanation:

A flat, horizontal line indicates a phase change.

The temperature does not increase because the added heat goes into converting one phase into another.

A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.

C is wrong. An upward-sloping line indicates that the temperature is increasing with time.

lisov135 [29]3 years ago

5 0

Answer:

horizontal line

Explanation:

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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

4 years ago

Be sure to show all 4 steps for each of the problems below.

Kisachek [45]

Answer:

1=920

2=2300

Explanation:

from the question,

M¹=23

H¹=4

g=10(constant)

using the formula

P. E,= mgh

P. E= 23×10×4=920

P. E=23×10×10=2300

7 0

4 years ago

A car of mass m=1000kg is traveling at speed v and brakes. The skid marks are 20m long and the coefficient of kinetic friction i

scoray [572]

Answer:

v = 14 m/s

= 31.3 mph

The answer would be the same if the mass of the car were 2000 kg

Explanation:

Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.

V^2 = v^2 + 2×a×Δx

Now V, the final velocity is zero as the car stops

0 = v^2 + 2×a×Δx

v^2 = -2×a×Δx

v =√-2×a×Δx .....*

Now applying Newton's Second Law

Fnet = m×a

-Fk = m×a

-μ×N = m×a

-μ×m×g = m×a (The mass cancels out)

a = -μ×g

Substituting the value of a back to equation *

v = √-2×(-μ×g)×Δx

v = √-2×(-0.5×9.8)×20

v = 14 m/s

Therefore the speed the car was travelling with v = 14 m/s

which is equal to 31.3 mph

Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.

8 0

3 years ago

What is the name of the process where an object is exposed to nuclear radiation

Murljashka [212]

Answer:

Exposing objects to beams of radiation is called irradiation

6 0

3 years ago

Covalent bonds tend to make atoms more stable by helping them

Lelechka [254]

Achieve a full outer shell

6 0

3 years ago