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2 years ago

To reach a distance of 2 km, a car has to cover a path of 2.4 km in 10 minutes. Calculate its average speed and velocity

Physics

1 answer:

nexus9112 [7]2 years ago

5 0

Answer:

Average speed = 0.24 km/min.

Average velocity = 0.2 km/min.

Explanation:

Finding average velocity :

Least possible distance / Time taken
2 km / 10 min
0.2 km/min.

Finding average speed :

Path distance / Time taken
2.4 km / 10 min
0.24 km/min.

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You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago

Please help with these questions

lord [1]

When do you gotta turn it in?

8 0

2 years ago

Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate

Helen [10]

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^{4}-T_{c}^{4}\right)$

So this law is application for calculating power absorbed by any surface.

4 0

3 years ago

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

E = 420.9 N/C

3 0

3 years ago

The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch

AnnZ [28]

Answer:

Explanation:

Given an LC circuit

Frequency of oscillation

f = 299 kHz = 299,000 Hz

AT t = 0 , the plate A has maximum positive charge

A. At t > 0, the plate again positive charge, the required time is

t =

t = 1 / f

t = 1 / 299,000

t = 0.00000334448 seconds

t = 3.34 × 10^-6 seconds

t = 3.34 μs

it will be maximum after integral cycle t' = 3.34•n μs

Where n = 1,2,3,4....

B. After every odd multiples of n, other plate will be maximum positive charge, at time equals

t" = ½(2n—1)•t

t'' = ½(2n—1) 3.34 μs

t" = (2n —1) 1.67 μs

where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

t''' = ½(2n—1) 1.67μs

t"' = (2n —1) 0.836 μs

where n = 1,2,3...

6 0

3 years ago