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3 years ago

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a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal . It remained in the air for 2 s. Whic

h of the following quantities was constant while the baseball was in the air?

1 answer:

liberstina [14]3 years ago

3 0

Answer:

Explanation:acceleration

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A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth? A. 37.93 N

Kryger [21]

Force of gravity = Mass x Gravitational force
Fg=mg
Fg = (5.7)(9.8)
Fg= 55.86 N

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A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W

dalvyx [7]

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

$F_g = mg$

now the acceleration of ball is given as

$a = \frac{F_g}{m}$

$a = \frac{mg}{m} = g$

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

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What is the most visited State in the USA.

Harrizon [31]

1-California
2-Florida
3-Texas
4-New York

So I’m sure that to most visited state in the USA is California

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One particle has a charge of 2.15 x 10^ -9 while another particle has a charge of 3.22 x 10^ -9 If the two particles are separat

marin [14]

Answer:

B. 2.77 x $10^{-4}$ N

Explanation:

The required force can be calculated by:

F = $\frac{Kq_{1}q_{2} }{d^{2} }$

Where F is the force between the particles, K is the coulomb's constant (9 x $10^{9}$ $Nm^{2}/C^{2}$ ), $q_{1}$ is the charge on the first particle, $q_{2}$ is the charge on the second particle and $d^{2}$ is the distance between the particles.

So that:

F = $\frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9} }{(0.015)^{2} }$

= $\frac{6.2307*10^{-8} }{(2.25*10^{-4} }$

= 2.7692 x $10^{-4}$

The force between the particles is 2.77 x $10^{-4}$ N.

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3 years ago

Two identical conducting spheres each having a radius of 0.500 cm are connected by a light, 1.90-m-long conducting wire. A charg

Ivan

Answer: 4.19 N

Explanation: In order to determinate the tension applied on the wire we have to calculate the electric force between the conductor spheres connected by the wire.

As the wire is a conductor the spheres are at same potential so we have:

V1=V2

V1=k*Q1/r1 and V2=k*Q2/r2

where r1=r2, then

Q1=Q2

so the electric force is given by:

F=k*Q^2/d^2 where d is the distance between the spheres.

Finally replacing the values, we have

F=9*10^9(41*10^-6)^2/(1.9)^2= 4.19 N

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