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3 years ago

On a cold winter day, the flow of heat is from the outside in. A.True B.False

Physics

2 answers:

fredd [130]3 years ago

8 0

Your answer s obviously false

lbvjy [14]3 years ago

7 0

Answer:

False

Explanation:

ooga booga, cold winter day cold outside, heat not outside ooga booga.

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A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force

SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, $\theta=30^{\circ}$

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

$F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N$

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0

3 years ago

I need hellp with thisss plssss????????!!!!

Leto [7]

Answer:

The graph appears to be in error.

The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5

The work done (F * S) is the area of the rhombus

1/2 * (5 +15) * 5 = 50 J

8 0

2 years ago

Using this formula Vj = V; + at, If a vehicle starts from rest

Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

$v_f=0+4.5*8$

$v_f=36\ m/s$

The final speed of the vehicle is 36 m/s

5 0

3 years ago

Which state of matter has atoms that are spread out and bouncy?

NikAS [45]

The stage where atoms are spread out and bouncy is the gas stage.

7 0

2 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago