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vaieri [72.5K]
3 years ago
9

im bored and lonely put those two together and youve got me. any guys wanna create a zo.om just for fun. im dyying of boredom.

Physics
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

just guys

Explanation:

and if not i need how old you are sorry just trying to be safe

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If a car skids 66 ft on wet concrete, how fast was it moving when the brakes were applied? (Round your answer to the nearest who
Sonbull [250]

Answer:

If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.

Explanation:

To determine how fast the car was moving, after skidding, the formula below is used:

V = √32*fd

V is the car's speed (ft/s)

d is skid length (ft) = 66 ft

f is the coefficient of friction determined by the material the car was skidding on.

Coefficient of friction for wet concrete is 0.65

V = √32*fd

V = √32 *0.65* 66

V = 242.679 ft/s ≅ 243 ft/s (nearest whole number)

If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.

8 0
4 years ago
Gerald works on a futuristic space station servicing interstellar space ships. He observes a ship traveling directly away from t
julsineya [31]

Answer:

9.14x10^22J

Explanation:

r= 1/√1-(0.17c)²/c²

1.01477

E= rmc²

9.14*10^22J

5 0
3 years ago
A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st
coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2  we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0  vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0
3 years ago
Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet
yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2)  and the remainder leaves at (3)

The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

m_{3}=m_{1}-m_{2}

m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

We need to calculate the inlet velocity

Using formula of velocity

v=\dfrac{m}{\rho A}

Put the value into the formula

v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

v=21.9\ m/s

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0
4 years ago
You pull a block of mass m across across a frictionless table with a constant force. you also pull with an equal constant force
KiRa [710]
For any mass m:

a = F/m
v = √2*F/m*s = √2F/sm = k/√m
Momentum = mv = k√m
Energy = 1/ mv² = 1/2 m.k²/m = 1/2k²

SO
Both will have same energy
The larger mass will have greater momentum
4 0
3 years ago
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