Someone please help me!

MY LAST ONE<,Brainliest for best answer!

grigory [225]

_Award brainliest if helped!
Mechanical Advantage = Force by Hammer / Force by Nail = 160/40 = 4

8 0

3 years ago

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is 55 mi/h

so if he drives all the to the meeting with max speed then it takes $=\frac{450}{55}=8.181 h$

and total allowable time is 10.8

Therefore longest time he can spend over dinner is $10.8-8.181 \approx 2.6 hours$

5 0

3 years ago

The critical angle between a medium and air is 430. What is the speed of light in that medium?

vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

= 1/sin(430)

= 1.06

then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

= (3×10^8)/(1.06)

= 281907786.2 m/s

Therefore, the speed of light is that medium is 281907786.2 m/s.

8 0

3 years ago

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0

3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

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