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andreev551 [17]
3 years ago
5

Starting from rest, a truck travels in a straight line for 8.0 s with a uniform acceleration of +1.6 m/s2. The driver then appli

es the brakes for 3.0 s, causing a uniform acceleration of −3.0 m/s2 over that time.(a) What is the truck's speed at the end of the braking period?
(b) What is the total distance traveled by the truck (from the point where it started at rest to the end of the braking period)?
Physics
1 answer:
lana [24]3 years ago
4 0
Because it should be the total number
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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

6 0
3 years ago
A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3
Alex73 [517]
Sorry I just need points
7 0
3 years ago
While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room an
Bogdan [553]

Answer:

the time needed for her to close the door is 1.36 s.

Explanation:

given information:

Force, F = 220 N

width, r = 1.40 m

weight, W = 790 N

height, h = 3.00 m

angle, θ = 90° = π/2

to find the times needed to close the door we can use the following equation

θ = ω₀t + 1/2 αt²

where

θ = angle

ω = angular velocity

α = angular acceleration

t = time

in this case, the angular velocity is zero. thus,

θ = 1/2 αt²

now, we can find the angular speed by using the torque formula

τ = I α

where

τ = torque

I = Inertia

we know that

τ = F r

and

I = 1/3 mr²

so,

τ = I α

F r = 1/3 mr² α

α = 3 F/mr

  = 3 F/(w/g)r

  = 3 (220)/(790/9.8) 1.4

  = 5.85 rad/s²

θ = 1/2 αt²

π/2 = 1/2 5.85 t²

t = 1.36 s

5 0
3 years ago
List two examples of when you would use light years as a measurement.
astra-53 [7]
One is when you are measuring a distance in space! I don't know the other but hope you find another example!
7 0
3 years ago
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