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Setler [38]
3 years ago
13

Bolt starts the race not moving, but then increased his speed until he reaches a top speed. Once Bolt reaches his top speed he m

aintains that speed for a few seconds to:
a) move at a constant velocity
b) accelerating by moving faster
c) accelerating by moving slower
Is it a, b, or c?
Physics
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

a) move at a constant velocity

Explanation:

You might be interested in
Which of the following represent units of an electric field? select all that apply
zimovet [89]

Answer:

Volts/Meter

Newtons/Coulomb

Explanation:

Volts/ Meter and Newtons/Coulomb both are same and the units of Electric field intensity or electric field strength.

Electric field strength E is the force per unit charge. It is measured in Newton/Coulomb in SI unit. It is a vector quantity directed in the direction of force.

Mathematically,

           Electric field strength = Force/Charge

                                           E = F / q₀

                                               = Newton / Coulomb = NC⁻¹             1

We know that

                         Newton = Joule/meter            2

Also

                           Volt = Joule/Coulomb             3

So put 3 in 2 we get

                          Newton = (Volt Coulomb)/meter       put in 1

   

                            E =  (Volt Coulomb)/(meter Coulomb)

                               = Volt / meter

Hence

                  Newton / Coulomb = Volt / meter    

                         

                                               

8 0
3 years ago
Read 2 more answers
A dog sits 2.6 m from the center of a merry- go-round. a) If the dog undergoes a 1.7 m/s^2 centripetal acceleration, what is the
alexgriva [62]

Answer:

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Explanation

Conceptual analysis :

The dog and the merry-go- round describes a circular motion, then, the following formulas apply :

a_{c} =\frac{v^{2} }{r} Formula (1)

v = ω *r   Formula (2)

Where:

a_{c} : Centripetal acceleration(m/s²)

v: linear speed or tangential (m/s)

r :  radius of the circle (m)

ω : angular speed ( rad/s)

Data

r= 2.6 m

a_{c} =  1.7 m/s²

Problem develpment

a) We replace data in the formula 1 to calculate the dog's linear speed(v):

a_{c} =\frac{v^{2} }{r}

1.7 =\frac{v^{2} }{2.6}

v^{2} =1.7*2.6 = 4.42

v=(\sqrt{4.42})\frac{m}{s}

v= 2.1 m/s

b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).

v = ω *r

2.1 = ω *2.6

ω = 2.1/2.6

ω = 0.807 rad/s

6 0
3 years ago
A film badge worn by a radiologist indicates that she has received an absorbed dose of 2. 5 x 10-6 gy. the mass of the radiologi
Anestetic [448]

Energy absorbed by the radiologist is 1.775*10^-6J.

To find the answer, we have to know about the radiation.

<h3>How much energy has she absorbed?</h3>
  • We have the expression for dose of absorption as,

           D=\frac{E}{m}

where; E is the energy absorbed and m is the mass of the body.

  • From the above expression, substituting appropriate values given in the question, we get,

         E=D*m=2.5*10^{-6}J/kg*71kg=1.775*10^{-4}J

Thus, we can conclude that, the Energy absorbed by the radiologist is 1.775*10^-6J.

<h3 />

Learn more about the radiation here:

brainly.com/question/24491547

#SPJ4

4 0
2 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
7. An 8 kg ball is travelling to the east at 10 ms', collides with a 2 kg ball travelling to the
Ymorist [56]

Answer:

The final velocity of the ball is 7m/s

Explanation:

M1=8kg,  V1 =10m/s , M2=2kg , V2=-5m/s

initial momentum before collison

m1v1+m2v2

=8×10 +2×(-5)  =80-10  = 70kg m/s

final momentum after collison

=(m1+m2)×v

=(8+2)×v

=10v

According to the law of conversion of momentum

initial momentum =final momentum

70=10v

10v=70

v=70/10

v=7m/s

3 0
3 years ago
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