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2 years ago

6

Which of toby's answers is a correct description of what happens when a 1-kg cart traveling at 1 m/s collides inelastically with

another 1-kg cart at rest?.

1 answer:

Nonamiya [84]2 years ago

4 0

Answer:

hi I don't know sorry sorry forgive me

Explanation:

sorry

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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

--

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

--

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

--

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago

A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

$Normal force = Gravitational force - centripetal force$

At the foot of a round hill

$Normal Force = centripetal force + Gravitational force$

4 0

3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

alekssr [168]

Answer:

The change in temperature is $\Delta T = 1795 K$

Explanation:

From the question we are told that

The temperature coefficient is $\alpha = 4 * 10^{-3 }\ k^{-1 }$

The resistance of the filament is mathematically represented as

$R = R_o [1 + \alpha \Delta T]$

Where $R_o$ is the initial resistance

Making the change in temperature the subject of the formula

$\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]$

Now from ohm law

$I = \frac{V}{R}$

This implies that current varies inversely with current so

$\frac{R}{R_o} = \frac{I_o}{I}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]$

From the question we are told that

$I = \frac{I_o}{8}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]$

=> $\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )$

$\Delta T = 1795 K$

6 0

3 years ago

A force of 5N compresses a spring by 4cm. Find the force constant of the spring.

marissa [1.9K]

Answer:

k = 1,250 N/m

Explanation:

Use the formula F=kx, with F=5N and x=0.04m

Then the spring constant (k) is 5/0.04

6 0

2 years ago

What is the magnitude of the gravitational force acting on the earth due to the sun?

expeople1 [14]

Answer: $3.524(10)^{22}N$

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m1=1.99(10)^{30}kg$ is the mass of the Sun

$m2=5.972(10)^{24}kg$ is the mass of the Earth

$r=1.50(10)^{11}m$ is the distance between the Sun and the Earth

Substituting the values in (1):

$F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}$ (2)

Finally:

$F=3.524(10)^{22}N$ This is the gravitational force acting on the earth due to the sun

3 0

2 years ago

Read 2 more answers