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bazaltina [42]
3 years ago
15

Can someone pls help, thank you in advance!

Physics
1 answer:
zzz [600]3 years ago
7 0

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

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Which factor is most important in determining climate?
alisha [4.7K]
Temperature. The other three dont have anything to do with determining climate

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KIM [24]

Answer:

Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.

Explanation:

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3 years ago
What is the power output of a pump which can raise 60kg of water height to height of 10m every minute
sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0
2 years ago
A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
Vesna [10]

Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

3 0
3 years ago
An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into
Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

3 0
3 years ago
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