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3 years ago

PLS HELP I NEED ANSWER HOMEWORK DUE SOON

Physics

1 answer:

vovangra [49]3 years ago

4 0

Answer:

I think its 1

Explanation:

The reason why I think it's 1 is because 0.5m/s^2 /0.5 m/s= 1 s

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A car traveling 20 m/s to the north collides with a car traveling 10 m/s to the

sergiy2304 [10]

Answer: C

Explanation:

In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.

Change in momentum of the system will be momentum after collision minus total momentum before collision.

Since momentum is a vector quantity, the direction will also be considered.

Momentum = MV - mU

Let

M = 800 kg is going north

at V = 20 m/s and the other car

m= 800 kg is going south

at U = 10m/s.

Substitute all the parameters into the formula

Momentum = (800 × 20) - (800 × 10)

= 8000 kgm/s

The final momentum after collision will also be equal to 8000 kgm/s

Change in momentum = 8000 - 8000

Change in momentum = 0

5 0

3 years ago

Read 2 more answers

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago

A toy race car travels down a 25cm ramp in 5 seconds. Calculate its speed. A. 125 cm/sec B. 5 cm/sec C. 0.2 cm/sec D. 20 cm/sec

kipiarov [429]

This would be B, 5cm/sec.
To get this answer you would need to divided 25 which is how large the ramp is, and 5 seconds, the amount of time it took to travel down the ramp.

7 0

3 years ago

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

olasank [31]

A) $P = 75 \times (2.7\times10^7) = 2.025\times10^9 W$

b) $Efficiency = \frac{8\times10^8}{2.025\times10^9} \times 100 \approx 39.5\%$

4 0

3 years ago

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Which number is not rounded correctly?

LekaFEV [45]

B: 210.8 rounded to 210 is totally wrong, and the reason why is because 210.8 rounded to the nearest whole number is 211, not 210. So B is the one with the error (this option is correct) and the other user that said D, is wrong since 18.42 does round to 18.4.

Hope this helped!

Nate

5 0

3 years ago

Read 2 more answers