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1 year ago

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A 2.0kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with a speed of 5.0 m

/s. What is the instantaneous speed of the point of the disk that makes contact with the surface?

1 answer:

Alexandra [31]1 year ago

4 0

Answer:

Instantaneous speed of contact point will be ZERO

Explanation:

As we know that disc is rolling without slipping on horizontal surface

So here the speed of center of the disc is given as

v = 5 m/s

now at the contact point the tangential speed will be in reverse direction

$v_t = R\omega$

now we know that net contact speed with respect to its lower surface must be zero

$v_{net} = v - v_t = 0$

so net velocity of contact point with respect to its lower surface must be ZERO here

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Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s

Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

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1 year ago

A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the

kakasveta [241]

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass = x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained by glass + Heat gained by water

Heat gained by glass = x × 0.18 × 20

Heat gained by water = 74 × 1.0 × 20

Hence;

2000 = (x × 0.18 × 20) + ( 74 × 1.0 × 20)

2000 - 1480 = (x × 0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.

Learn more: brainly.com/question/1446583

3 0

1 year ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

1 year ago

Read 2 more answers

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

gregori [183]

The answers are:

a) $Work=125,923.61J$

b) $Power=1.46watt$

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

$1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}$

Also, we need to find the mass given the density of the blood.

$1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg$

Now, calculating how much work does the heart do in a day, we have:

$Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J$

Then, calculating what is the power output and its horsepower, we have:

$Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt$

Have a nice day!

7 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago