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3 years ago

In Newtons first law of motion how is an object at rest affected

Physics

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Explanation:

Meaning that if an object is just sitting somewhere unless something else causes it to move it's not going anywhere.

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An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav

fiasKO [112]

Answer:

θ = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

p₀ = m v₀

Final after the crash before starting the movement

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v

v = v₀ m / (m + M)

v = 2.0 0.015 / (0.015 +0.080)

v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

Em₀ = K = ½ (m + M) v²

Higher

$E_{mf}$ = U = (m + M) g y

Em₀ = $E_{mf}$

½ (m + M) v² = (m + M) g y

y = ½ v² / g

y = ½ 0.316² / 9.8

y = 0.00509 m

Let's look for the angle the height from the pivot point is

L = 0.40 / 2 = 0.20 cm

The distance that went up is

y = L - L cos θ

cos θ = (L-y) / L

θ = cos⁻¹ (L-y) / L

θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

θ = 12.95º

4 0

3 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$

which is equivalent to the gradient of the line in the velocity-time graph.

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

$a=-1 m/s^2$ is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

$T=-\frac{u}{a}=-\frac{10}{-1}=10 s$

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

$a=-1 m/s^2$ is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

$s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m$

7 0

3 years ago

You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist

omeli [17]

The acceleration of the ball after leaving the hand is $9.8 m/s^2$ downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

$F=mg$

where m is the mass of the ball and g is the acceleration of gravity ( $g=9.8 m/s^2$ ), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

$a=\frac{\sum F}{m}$

where

$\sum F$ is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

$a=\frac{mg}{m}=g=9.8 m/s^2$

This means that the acceleration of the ball remains $9.8 m/s^2$ downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

The shanghai maglev train is capable of reaching speeds of up to 350km h-1 what is the speed in m/h

LenaWriter [7]

Answer:

The shanghai maglev train is capable of reaching speeds of up to 217.48 in mph

217.48 mph

Explanation:

8 0

3 years ago