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faust18 [17]
3 years ago
15

A ball is thrown upward at time t=0 from the ground with an initial velocity of 60 m/s (~ 134 mph). Assume that g = 10 m/s2.At w

hat time t does the ball reach its apex (highest point)?
Physics
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

6 second

Explanation:

initial velocity of ball, u = 60 m/s

g = 10 m/s^2

Let the ball takes time t to reach at the maximum height

We know that at maximum height, the velocity of ball is zero.

v = 0 m/s

Use first equation of motion

v = u + gt

0 = 60 - 10 x t

t = 6 second

Thus, the ball takes 6 second to reach to maximum height.

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I need the answer asap everyone have a good day bye
Slav-nsk [51]

Im pretty sure its A cuz is closer to the earth.

5 0
3 years ago
A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

8 0
3 years ago
Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

5 0
3 years ago
What happens to particles of the medium in a mechanical wave?
Karo-lina-s [1.5K]
They flow i guess but dont quote me on that
3 0
3 years ago
Read 2 more answers
Using the equation E = mc2, calculate how many joules of energy would be produced by converting 6.8 × 10-6 kg of matter into ene
mixer [17]

Answer:

Explanation:

Using the formula : E = mc²

Where m = mass = 6.8 × 10^-6 kg

c = speed of light = 3 ×10^8 m/s

The amount of joules of energy that would be produced equals :

Plugging in our values :

E = mc²

E = (6.8 × 10^-6) kg × (3 ×10^8)²m/s

E = (6.8 × 10^-6) kg × 9 × 10^16 m/s

E = 61.2 × 10^(-6 + 16)

E = 61.2 × 10^10 J

6 0
3 years ago
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