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2 years ago

A car is moving with an initial relocity of

Physics

1 answer:

MA_775_DIABLO [31]2 years ago

4 0

Answer:

The final acceleration of the car, v = 70 m/s

Explanation:

Given,

The initial velocity of the car, u = 20 m/s

The acceleration of the car, a = 10 m/s²

The time period of travel, t = 5 s

Using the I equations of motion

v = u + at

= 20 + 10(5)

= 20 + 50

= 70 m/s

Hence, the final acceleration of the car, v = 70 m/s

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The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0

2 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago

In nature, where do fusion reactions take place?

hjlf

Fusion reactions happen in stars

3 0

2 years ago

Read 2 more answers

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611$

Converting to SI units

$10\ ft/s=10\times \dfrac{1}{3.281}$

$32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}$

$2\ ft=2\times \dfrac{1}{3.281}$

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611$

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

7 0

2 years ago

Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t

MrRa [10]

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

$EMF = -\frac{\delta \phi _B}{\delta t}$

$(\frac{\delta \phi _B}{\delta t} )$ is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = $\frac{\delta \phi _B}{\delta t}$ = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

8 0

3 years ago