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Anna [14]
3 years ago
15

When does a ball that is thrown vertically upwards reach its maximum speed?

Physics
1 answer:
ladessa [460]3 years ago
4 0
As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
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A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has
LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

<h3>What is the distance?</h3>
  • The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
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(c) the distance D where the object comes to rest.

W_{total} =ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1*2.3^{2}

⇒D=\frac{0.5*2.3^{2} }{2.45}

⇒1.08.m

To learn more about distance, refer to:

brainly.com/question/4998732

#SPJ4

5 0
1 year ago
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet
enot [183]

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  E_o   =  611\  V/m

   

Generally the  magnetic  field amplitude is  mathematically represented as

              B_o  =  \frac{E_o }{c }

Where c is the speed of light with a constant value

         c = 3.0 *0^{8} \ m/s

So  

        B_o   =  \frac{611 }{3.0*10^{8}}

         B_o   =  2.0 4 *10^{-6} \  Vm^{-2} s

Since 1  T  is  equivalent to  V  m^{-2} \cdot  s

         B_o  =  2.0 4 *10^{-6} \ T

6 0
3 years ago
One astronomer believes that the density of the universe remains constant. One physicist believes that the density of the univer
Llana [10]
I think the third option is the answer 
6 0
3 years ago
Read 2 more answers
Calculate the mechanical advantage of a hammer, if the input force is 125 N
olganol [36]

Answer:

16

Explanation:

Mechanical advantage = force out / force in

MA = 2000 N / 125 N

MA = 16

5 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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