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2 years ago

7

John and Caroline go out for a walk one day. This graph represents their distance from home.Which statement accurately describes

their walk?
They walked uphill quickly, then they were still, then they walked downhill slowly.
They walked faster and faster until they walked at a constant rate, then they walked slower and slower.
They walked quickly, then they were still, then they walked slowly.
They walked in one direction, then they were still, then they walked in the opposite direction.

2 answers:

kotegsom [21]2 years ago

7 0

I get where they need to go

Zepler [3.9K]2 years ago

3 0

Answer:

C) They walked in one direction, then they were still, then they walked in the opposite direction.

Explanation:

yw.

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A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis

CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car $v=10 m/s$

mass of clump $m=0.5 kg$

Radius of car tire $r=0.2 m$

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

$F_c=\frac{mv^2}{r}$

where $m=mass\ of\ element$

$v=speed$

$r=distance\ from\ center$

$F_c=\frac{0.5\times 10^2}{0.2}$

$F_c=250\ N$ (inward)

3 0

3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$

The compression of the net is 1.52 m

4 0

3 years ago

Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source

kodGreya [7K]

Resistance can be calculated by Ohm's law

As per ohm's law we will have

$V = i * R$

here we will have

voltage = 220 volts

current = 10 A

So by the above formula we will have

$220 = 10* R$

$R = 22 ohm$

So resistance of the bulb is 22 ohm.

4 0

3 years ago

A wheel at rest gains an angular momentum of 10.7kgm²/s in 8s. if the moment if inertia of the wheel is 2.2kgm², how many revolu

EleoNora [17]

Answer:

Revolutions made before attaining angular velocity of 30 rad/s:

θ = 3.92 revolutions

Explanation:

Given that:

L(final) = 10.7 kgm²/s

L(initial) = 0

time = 8s

<h3>Find Torque:</h3>

Torque is the rate of change of angular momentum:

$T = \frac{L(final)-L(initial)}{t}\\T = \frac{10.7-0}{8}\\T=1.34 Nm$

<h3>Find Angular Acceleration:</h3>

We know that

T = Iα

α = T/I

where I = moment of inertia = 2.2kgm²

α = 1.34/2.2

α = 0.61 rad/s²

<h3>Find Time 't'</h3>

We know that angular equation of motion is:

ω²(final) = ω²(initial) +2αθ

(30 rad/s)² = 0 + 2(0.61 rad/s²)θ

θ = (30 rad/s)²/ 2(0.61 rad/s²)

θ = 24.6 radians

Convert it into revolutions:

θ = 24.6/ 2π

θ = 3.92 revolutions

3 0

2 years ago

In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite

Eddi Din [679]

Answer:

option C

Explanation:

the ball is moving circular around the pole

Angular momentum of the system is constant

J = I ω

now,

$\omega\ \alpha\ \dfrac{1}{I}$

$\omega\ \alpha\ \dfrac{1}{mr^2}$

$\omega\ \alpha\ \dfrac{1}{r^2}$

the rope radius is decreasing as it revolving around the pole

angular speed is inversely proportional to radius.

so, the angular speed will increase.

The correct answer is option C

7 0

2 years ago