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Arada [10]
3 years ago
5

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati

on of -3.9 m/s/s, determine the speed of the Jaguar before it began to skid to a stop.

Physics
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:  47.6 m/s

Explanation:  Please see attached for the calculation and formula.

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You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s
BlackZzzverrR [31]
1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>
6 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
3 years ago
Two 10cm diameter metal disks separated by a 0.63mm thick piece of pyrex glass are charged to a potential difference of 1000V. D
inessss [21]
Parallel-plate capacitor has there fore formula is

<span>C=(<span>ϵ0</span>A)/d
putting values</span>C=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F then Q=CV=1.1*10^-10*1000=1.1*10^-7C 
as
<span>η=Q/A</span><span>therefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
hope this helps</span>
5 0
3 years ago
7.
kiruha [24]

Answer:

The given figure shows two men M and N facing two flat and hard walls, wall 1 and wall 2. Man N fires a gun. Man M hears two echoes, one from wall 1 and second from wall 2. The speed of sound in

air is given to be 325 m/s. After the firing of the gun by the man N, the man M will hear the first echo in how many seconds?

5 0
2 years ago
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