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Nady [450]
3 years ago
6

A soccer player is running upfield at 10m/s and comes to a stop in 3 seconds facing the same direction. What is his acceleration

?
Physics
2 answers:
Sonja [21]3 years ago
5 0

Answer : His acceleration is, -3.3m/s^2

Explanation :

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 0 m/s

u = initial velocity  = 10 m/s

t = time = 3 s

a = acceleration = ?

Now put all the given values in the above equation 1, we get:

0m/s=10m/s+a\times (3s)

a=-3.3m/s^2

Therefore, his acceleration is, -3.3m/s^2

Mekhanik [1.2K]3 years ago
4 0
The players acceleration is 3.33 m/s/s

Acceleration= Velocity/Time

A =10/3
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A plane is flying 2400 miles from a to
geniusboy [140]
From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560

let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs

(2400-640t) / 640 = 640t / 560
560(2400 - 640t) = 640t x 640
t = 1.75hrs
5 0
3 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
3 years ago
What is a rubens tube
Free_Kalibri [48]

Answer:

its an antique physics apparatus for demonstrating acoustic standing waves in a tube.

6 0
3 years ago
50 POINTS
sweet-ann [11.9K]

Answer:

The acceleration is constant and positive

Explanation:

The straight line indicates that the acceleration is constant, while the positive slope indicates that the line is positive.

7 0
3 years ago
Read 2 more answers
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