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Anna11 [10]
3 years ago
5

A car with mass m travels over a hill with a radius of curvature of r at a speed of 15 m/s. What is the normal force on the car

when the car is at the top of the hill?
Physics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

zero

Explanation:

The computation of the normal force is shown below:

As we know that

F_c = mg - N

F_c = mv^2 ÷ r

N = mg - mv^2 ÷ r

N = m(g - v^2 ÷ r)

Assume that

The mass of the car is 1200 kg

And, r = 10 m

So,

= 1200 (9.8 - 15^2 ÷ 10)

= -15240 N

Since it comes in negative so the normal force would be zero

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A machine does 500 j of work in 20 sec. What is the power of this machine?
NNADVOKAT [17]

Explanation:

P=W/t

P=500/20

P=25 W

6 0
3 years ago
Help please
Anarel [89]

Answer:

1) D, 2) D, 3) B, 4) B, 5) C

Explanation:

You are asked to select the correct answer

1)  The conservation of energy is one of the most important principles of physics that allows solving countless problems in life.

the correct answer is D

2) when a body falls, the gravitational potential energy is transformed into kinetic energy and both are transformed into thermal energy

the correct answer is D

3) When the gravitational potential energy is maximum, the kinetic energy is minimum and when the kinetic energy is maximum, the gravitational energy is minimum.

Correct answer B

4) speed is defined by

         v = x / t

so the correct answer is B in the SI system

5) when we repeat a measurement several times, the random or statistical errors decrease, therefore the confidence of the measurement increases.

The correct answer is C

4 0
3 years ago
An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for
VikaD [51]
We know that
g = LcosΘ 
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
8 0
3 years ago
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A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni
Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A

That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

7 0
3 years ago
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
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