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hoa [83]
3 years ago
10

A 905 kg test car travels around a 3.25 km circular track. if the magnitude of the force that maintains the car's circular motio

n is 2140 N, what is the car's tangential speed?
Physics
1 answer:
Verizon [17]3 years ago
8 0
Link provided in other answer is a SCAM, don’t click on the link !!!!
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Dinah is playing cards with her friends, and it is her turn to deal. A card with a mass of 2.3 g slides 0.35 m along the table b
Inga [223]

Answer:

soln,

=2.3+0.35-0.24

=2.41 answer

4 0
3 years ago
Phosphorus (P) is an element with an atomic number of 15 and an atomic mass of 31. How many neutrons are in an atom of phosphoru
BaLLatris [955]

(31-15 = 16).

Explanation:

the element phosphorus (P) has an atomic number of 15 and a mass number of 31. Therefore, an atom of phosphorus has 15 protons, 15 electrons, and 16 neutrons

5 0
3 years ago
A 800 J <br> B -800 J <br> C 400 J <br> D -400 J <br> ??
Ierofanga [76]

I think it will be 800j because right absorb mean getting more

8 0
4 years ago
When exiting the highway, a 1100-kg car is traveling at 22 m/s. The car's kinetic energy decreases by 1.4×105J The exit's speed
Vesna [10]
I think you want to determine the exit speed?

You have to determine how much velocity was decreased by calculating it from the kinetic energy.

KE = (1/2)mv²
1.4 x 10^5 = (1/2)*(1100)v²
v² = 254.55
v =15.95 m/s

So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.

So, the final speed was 6.05 m/s.

I hope I was able to help :)
7 0
3 years ago
Read 2 more answers
A research submarine has a 30-cm-diameter window that is 8.6cm thick. The manufacturer says the window can withstand forces up t
Triss [41]
What is missing is the density of salt water. Let's denote it ρ. 
<span>The pressure at depth d then is: </span>
<span>p = p0 + ρ g d </span>
<span>, where p0 is the atmospheric pressure at the surface of the sea. We assume it is equal to the 1 atm inside the sub. </span>

<span>Hence the pressure difference between inside and outside of the window is: </span>

<span>Δp = ρ g d </span>

<span>The force on the window is this pressure difference, multiplied by the area of the circular window of diameter D. The latter equals </span>
<span>A = ¼ π D². </span>

<span>So </span>
<span>F = p A = ¼ π D ²ρ g d. </span>
<span>This should be less than Fmax. </span>

<span>Hence d < Fmax / (¼ π D ²ρ g ). </span>
6 0
3 years ago
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