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Stells [14]
4 years ago
6

A juggler throws two balls up to the same height so that they pass each other halfway up when A is rising and B is descending. I

gnore air resistance and buoyant forces. Which statement is true of the two balls at that point?a. The only force acting on each ball is the gravitational force. b. There is an residual upward force from the hand on each ball. c. Only gravity acts on B but there is an additional residual force from the hand on A. d. There is a greater residual force from the hand on A than there is on B. e. There is an additional downwards force besides gravity on each ball.
Physics
1 answer:
madam [21]4 years ago
4 0

Answer:

<em>Only gravity acts on B but there is an additional residual force from the hand on A.</em>

Explanation:

All bodies are constantly under the effect of gravity. Gravity is what gives us weight here on earth. Gravity acts downwards, and helps to decrease the deceleration of a body moving up and accelerating a body that travels downwards. For the ball A, traveling upwards, the upwards movement is due to the force on it impacted on it from the hand. As A tries to go up, gravity tries to decelerate it until it will come to a stop and then fall downwards under gravity. For body B, descending down means that only gravity forces acts on it at that point, if we ignore buoyant forces and air resistance. And B accelerates as it falls down towards the juggler.

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An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
ivolga24 [154]

Answer:

12.24 m/s

Explanation:

Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.

Using the formula,

a = v/t................ Equation 1

Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.

making v the subject of the equation,

v = at ................. Equation 2

Given: a = 5.1 m/s², t = 2.4 s.

Substitute into equation 2

v = 5.1(2.4)

v = 12.24 m/s.

Hence, the speed of the sprinter = 12.24 m/s

3 0
3 years ago
A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad
OverLord2011 [107]

Answer:

Explanation:

Given

radius of  path r=2\ m

Velocity of Particle \theta =5t^2 rad

where t=time in seconds

angular velocity of particle is given by

\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}

\omega =2\times 5t=10\cdot t

And angular acceleration is given by

\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}

\alpha =10 rad/s^2

tangential acceleration is a_t=\alpha \times r

a_t=10\times 2=20\ m/s^2

Centripetal acceleration a_c=\omega ^2\times r

a_c=(10t)^2\times 2=200t^2

net acceleration is sum of tangential and centripetal force at any time t is given by

a_{net}=\sqrt{(a_c)^2+(a_t)^2}

a_{net}=\sqrt{(200t)^2+(20)^2}

a_{net}=20\sqrt{(10t)^2+1}\ m/s

                 

8 0
3 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
A TV is rated at 0.1 kW. This TV is used 3 hours per day. How much energy does the TV use per month (in kWh)?
exis [7]

Answer:

<em><u>A.9</u></em><em><u> </u></em><em><u>kWh</u></em>

Explanation:

Tv is rated 0.1 kW

in one day energy used will be 0.1kW ×3hr=0.3kwh

in 30 days energy used will be 0.3kwh×30=9kwh

6 0
4 years ago
A bus is traveling 25.0 km in 2 hours. What is the speed of the bus in millimeters per second?
trasher [3.6K]

Answer:

3472.222

Explanation:

In one hour, the bus travels at a 12.5 km. The speed is 12.5km/hr

Convert it to mm/s

And you get 3472.2mm/s

8 0
3 years ago
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