All categories

2 years ago

A truck is parked outside. When the sun goes down, the temperature

Physics

2 answers:

fomenos2 years ago

6 0

Answer:

it will get colder

Explanation:

Luda [366]2 years ago

3 0

Answer:

Colder.

When the temperature is high, the air in your tires takes up more volume, where as when the temperature is cold, the air takes up less volume.

You might be interested in

The box plots show the summer temperatures, in degrees Fahrenheit, in two cities. Summer Temperatures in City A Summer Temperatu

Inga [223]

Answer:b

Explanation:

7 0

3 years ago

Read 2 more answers

How do you find motion

saveliy_v [14]

Answer:

idk you got me there

Explanation:

8 0

3 years ago

How many joules are required to heat .250 kg of liquid water from 0 °C to 100 °C ?

dedylja [7]

Answer:

Specific heat of water is 4.186 J/g/C. The heat required to raise the temperature by

Here is mass of water being heated, specific heat of water and change in temperature.

Here .

Heat energy required is

Explanation:

4 0

2 years ago

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0

3 years ago

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

4 0

3 years ago

Read 2 more answers