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4 years ago

13

A car drives at a constant speed of 14 meters per second around a circle with a radius of 25 meters what is the centripetal acce

leration of of the car

1 answer:

mojhsa [17]4 years ago

4 0

Answer:

Cₐ=7.84ms⁻²

Explanation:

v=14ms⁻¹ r=25m

Cₐ=v²/r

Cₐ=14²/25

Cₐ=196/25

Cₐ=7.84ms⁻²

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Voltage needed to raise current to 3.75a using 20,20,200 resistor set

Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

We are given:

$R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega$

Using above equation, we get:

$\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega$

Calculating the voltage by using Ohm's law:

$V=IR$ .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = $9.52\Omega$

Putting values in equation 1, we get:

$V=3.75\times 9.52\\\\V=35.7V$

Hence, the voltage needed is 35.7 V

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3 years ago

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A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The s

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Answer:

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4 years ago

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

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$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

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3 years ago

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Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o

ella [17]

Answer:

$4.33\cdot 10^{-6}C$ , charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

$F=+0.30 N$ is the force

$r=75 cm=0.75 m$ is the distance between the spheres

$q_1 =q_2 =q$ because the two spheres have identical charge

Solving the formula for q, we find

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C$

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

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3 years ago

Light of wavelength 441.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen

pshichka [43]

Answer: R = 0.131 m

Explanation: The formulae for the distance between a fringe and the first (center) is given by

y = R×mλ/d

Where y = distance between first and nth fringe = 4mm = 4×10^-3m

λ = wavelength of light = 441.1nm = 441.1×10^-9m

R = distance between slits and screen =?

d = distance between slits = 0.29mm = 2.9×10^-4m

4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5

Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9

R = 1.16 × 10^-7/8.828×10^-7

R = 1.16/8.828

R = 0.131 m

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4 years ago