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blagie [28]
3 years ago
12

a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip

etal force on the car is 6000 newtons. what is the radius of the curve?
Physics
1 answer:
qaws [65]3 years ago
5 0
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity.  The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

         Force  =  (mass of the object) · (speed)² / (radius of the circle)

         F  =  m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

                                          F     =  m s² / r

Multiply each side by 'r':       F· r  =  m · s²

Divide each side by 'F':            r  =  m · s² / F    

We know all the numbers on the right side,
so we can pluggum in:

                      r  =       m       ·        s²      /     F

                      r  =  (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

                                      


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emmasim [6.3K]
Something hot like a fire , an eye of a stove , and the sun
4 0
4 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
4 years ago
An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har
sineoko [7]

Answer:

Acceleration will be equal to 2m/sec^2      

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to F=ma.....eqn 1

By hook's law spring force is equal to F=kx .....eqn 2

From equation 1 and equation 2

ma=kx

0.4\times a=8\times 0.1

a=2m/sec^2

So acceleration will be equal to 2m/sec^2

8 0
4 years ago
6. A 2.2 kg rock rests on the edge of a bridge that is 3.3 m above a river. What is the
vovikov84 [41]

Answer: 71 J

Explanation: The formula for gravitational potential energy is Epg= mgh.

So you plug in 2.2 as m, 3.3 as h and 9.81 as g. Once you do that you should get an answer of 71.2206. But because there are only two significant digits in the question, your answer must also have two significant digits.

6 0
4 years ago
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