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2 years ago

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A baseball is thrown horizontally at 55m/s. The ball slows down at a rate of -10 m/s2. How long is the ball in the air before co

ming to rest?

1 answer:

sergejj [24]2 years ago

6 0

The time of motion of the ball before coming to rest is determined as 5.5 s.

<h3>Time of motion of the ball</h3>

The time of motion of the ball before coming to rest is calculated as follows;

v = u + at

when the ball comes to rest, v = 0

0 = u + at

0 = 55 + (-10)t

0 = 55 - 10t

10t = 55

t = 5.5s

Thus, the time of motion of the ball before coming to rest is 5.5 s.

Learn more about time here: brainly.com/question/4931057

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Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

2 years ago

Read 2 more answers

The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4

scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

v = v₀ + a t

y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing.

Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s

v = v₀ + a t

$a = \frac{v-v_o}{\Delta t}$

$a = \frac{2-0}{8-0}$

a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

$a = \frac{6-2}{ 14-10}$

a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

$d_{total } = d_1 +d_2 + d_3 +d_4$

We look for every displacement.

d₁ = v₀ + ½ a₁ Δt²

d₁ = 0 + ½ 0.25 8²

d₁ = 8 m

In the second section the velocity is constant

d₂ = v₂ Δt₂

d₂ = 2 (10-8)

d₂ = 4 m

The third section.

d₃ = v₀ + ½ a t²

d₃ = 2 + ½ 1 (14-10) ²

d₃ = 10 m

The distance of the fourth section.

we look for acceleration

a₄ = $\frac{v-v_o}{\Delta t}$

a₄ = $\frac{0-6}{18-14}$

a₄ = -1.5 m / s²

d₄ = 6 + ½ (-1.5) (1814) ²

d₄ = -6 m

The total displacement is;

$d_{total}$ = 8 + 4 + 10 -6

$d_{total}$ = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0

2 years ago

You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t

shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0

3 years ago

Which ingredient would a company that makes super sour candy most likely use in a large quantity

stiks02 [169]

Suger........................

4 0

3 years ago

Read 2 more answers

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago