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3 years ago

The slope of the line on a speed-time graph tells the speed , true or false ?

Physics

2 answers:

KengaRu [80]3 years ago

8 0

True I hope this helps you out

shusha [124]3 years ago

4 0

This is true, the slope of the line of a speed graph depending on how high or low would make up for the speed of the line or the object the line is representing.

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An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce

notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0

3 years ago

A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th

kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

$P_1, V_1, T_1 & P_2, V_2, T_2$

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

$-Q=W+U_1----1$

$Q=-W+U_2-----2$

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

$U_1 & U_2$ change in internal Energy of gas

Thus from 1 & 2 we can say that

$U_1+U_2=0$

$n_1c_v(T-T_1)+n_2c_v(T-T_2)=0$

$T(n_1+n_2)=n_1T_1+n_2T_2$

$T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}$

where $n_1=\frac{P_1V_1}{RT_1}$

$n_2=\frac{P_2V_2}{RT_2}$

$T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

$T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

and $P=\frac{P_1V_1+P_2V_2}{V_1+V_2}$

6 0

3 years ago

________ is a force acting through distance.

Andreas93 [3]

High school???
No way
It's work.

8 0

3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$