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xxTIMURxx [149]
3 years ago
6

I really need help with this question someone plz help !

Physics
1 answer:
Vika [28.1K]3 years ago
7 0

Answer:

The answer is option 2.

Explanation:

Both sides are pulling the rope with equal force where the rope doesn't move. So they have a <u>b</u><u>a</u><u>l</u><u>a</u><u>n</u><u>c</u><u>e</u><u>d</u><u> </u><u>f</u><u>o</u><u>r</u><u>c</u><u>e</u><u>s</u>.

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You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha
Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

PE = mgh

At half height, this potential energy is

PE = \frac{1}{2}mgh

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

mgh = \frac{1}{2}mgh +KE

KE = \frac{1}{2}mgh = K/2

5 0
3 years ago
A sample of oxygen gas at 25.0°c has its pressure tripled while its volume is halved. What is the final temperature of the gas?
Marat540 [252]

Answer:

447 K

Explanation:

25 C = 25 + 273 = 298 K

Assuming ideal gas, we can apply the ideal gas law

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}

Since pressure is tripled, then P_2 / P_1 = 3. Volume is halved, then V_2 / V_1 = 0.5

T_2 = 298*3*0.5 = 447 K

4 0
3 years ago
Read 2 more answers
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
3 years ago
The closest star to our solar system is Alpha Centauri, which is 4.12 × 10^16 m away. How long would it take light from Alpha Ce
bonufazy [111]

Time = (distance) / (speed)

Time = (4.12x10^16 m) / (3 x10^8 m/s)

Time = 1.37 x 10^8 seconds

Divide the seconds by 86,400 to get days. Then divide the days by 365 to get years.

Time = about 4.35 years

4 0
3 years ago
Need help 8th grade science test need help
jolli1 [7]
ANSWER: d, average speed
7 0
2 years ago
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