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2 years ago

A car travels south at 30 m/s for 5 minutes. What is its velocity

Physics

2 answers:

Marysya12 [62]2 years ago

4 0

Miles per second right?
Hope that helped

MAVERICK [17]2 years ago

4 0

Answer:

Explanation:

Velocity = 30 (-j) m/s

Displacement = velocity x time = 30 x 5 x 60 = 9000 m South = - 9000 j m

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What factor causes atmospheric pressure

bagirrra123 [75]

Atmospheric pressure is caused by the weight of the atmosphere pushing down on itself and on the surface below it.
Pressure is defined as the force acting on an object divided by the area upon witch the force is acting.

7 0

2 years ago

Read 2 more answers

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

3 0

2 years ago

A Cambra pouce car traveling at 28 m/s slow

natka813 [3]

Answer:

t = 7.8 seconds

Explanation:

Given that,

The initial speed of the car, u = 28 m/s

Acceleration of the car, a = 3.6 m/s²

We need to find the time taken for the police car to come to Stop. When it stops, its final speed is equal to 0. So, using the equation of kinematics to find it i.e.

$v=u+at\\\\0=28+3.6t\\\\t=\dfrac{28}{3.6}\\\\t=7.8\ s$

So, the required time is 7.8 seconds.

7 0

2 years ago

Calculate the tension on the rope. A. 57.89 N B. 32.73 N C. 69.84 N D. 12.55 N

Ratling [72]

Hi there!

$\large\boxed{\text{B. 32.73N}}$

To calculate the tension, we must calculate the acceleration of the system.

Begin with a summation of forces:

∑F = -M₁gsinФ + T - T + M₂g

Simplify and solve for acceleration: (Tensions cancel out)

$a = \frac{-M_1gsin\theta + T - T + M_2g}{M_1+M_2}$

Plug in values. Let g = 10 m/s²

$a = \frac{-3(10)(sin30)+8(10)}{3+8} = 5.91 m/s^{2}$

Now, to find tension, let's sum up the forces acting on ONE block. For simplicity, we can look at the hanging block:

∑F = -T + W

ma = -T + W

Rearrange to solve for T:

T = W - ma

We know the acceleration, so plug in the values:

T = (8)(10) - (8)(5.91) = 32.73 N

3 0

2 years ago

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a

xxTIMURxx [149]

Answers:

a) $F_{g}=735 N$ and $n=732.47 N$ , hence $F_{g} > n$

b) $n_{poles}=735 N$ $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the centripetal force $F_{c}$ and the gravitational force $F_{g}$ (also called true weight) are directed "downward", while the normal force $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation: