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yan [13]
3 years ago
6

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the

hand is 2.30 m above the ground. How long is the ball in the air before it hits the ground?
Physics
1 answer:
My name is Ann [436]3 years ago
7 0

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

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A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
Liula [17]

The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

r=0.800 m-0.200 m=0.6 m

Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.


3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C

and the sign tells us that the field is directed toward negative x-direction.

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3 years ago
I need your opinion on this
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Conduction transfers Thermal energy by
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Answer:

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4 years ago
A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

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We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

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