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3 years ago

Y'all I need this comon please give me a right answer

Physics

1 answer:

Natali [406]3 years ago

5 0

It will be a virtual image that appears on the left side of the mirror

i hope this helps!

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A boy runs 400m at an average speed of 4.0m/s he runs the first 200m in 40 s how long does he take to run the second 200m?

IRISSAK [1]

If he runs at the same speed he will cover next 200m in 40s
that is at the average of 4.0m

8 0

3 years ago

The value of 1.0004 to the power 1 by 2 using Binomial approximation is

IgorLugansk [536]

Given:

The given value is $(1.0004)^{\frac{1}{2}}$ .

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

$(1.0004)^{\frac{1}{2}}$

It can be written as:

$(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}$

$(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004$ $[\because (1+x)^n=1+nx]$

$(1.0004)^{\frac{1}{2}}=1+0.0002$

$(1.0004)^{\frac{1}{2}}=1.0002$

Therefore, the approximate value of the given expression is 1.0002.

3 0

2 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago

A charge of 5.0 coulombs moves through a circuit in 0.50 second. What is the current in the circuit

user100 [1]

Answer:

Explanation:

i = 5/.5 = 10 Amps. Hope this helps :)

6 0

3 years ago

In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kozerog [31]

Answer: vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s vf2 = 1.5 m/s

8 0

3 years ago