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EleoNora [17]
3 years ago
6

Please need help on this

Physics
2 answers:
anyanavicka [17]3 years ago
7 0
Circuit breaker
I think is the answer
almond37 [142]3 years ago
4 0
NEED HELP WITH WHAT DONT SEEQUESTION
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• Usain Bolt can run at 10 m/s. If he runs for about 20 seconds, how far around the race track did he go?
yanalaym [24]

Answer:

200metters

Explanation:

because in one second hes going 10 metter in 20 second he will go 20×10=200

4 0
3 years ago
A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid
DedPeter [7]

Answer: IM 95%sure that the answer is B jus took the test got the answer right

Explanation:

6 0
3 years ago
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If the IMA of a machine is 2 and the effort force is 50 newtons, then the force applied to the resistance is _____. 48 N 50 N 25
zysi [14]

IMA stands for ideal mechanical advantage, which is the theoretical force amplification factor on an ideal mechanical device free of friction, deformations, etc.


If the applied force (effort) is 50N, then the force applied to the resistance is multiplied by the IMA=2 to get 100N.

7 0
3 years ago
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Please help me .... ​
miss Akunina [59]

Explanation:

Orbital speed= 2pi x radius / time period

=2pi x 1.5x10^11 / 365.25

=2.58x10^9m/day

8 0
3 years ago
Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug
Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

3 0
3 years ago
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