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2 years ago

Drag the tiles to the correct boxes to complete the pairs.

Physics

1 answer:

Brut [27]2 years ago

3 0

Answer:

5 percent = normal matter

68 percent = dark energy

27 percent = dark matter

Explanation:

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Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

3 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

8 0

3 years ago

HELP Giving all the points I can. Physics virtual lab report: circuit design<br><br> Please help me.

mafiozo [28]

Answer:text me I can help

Explanation:

5 0

3 years ago

Read 2 more answers

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th

NeX [460]

Answer:

3.14946 rad/s

Explanation:

$I_i$ = Intial moment of inertia

$I_f$ = Final moment of inertia

$\omega_i$ = Initial angular velocity

$\omega_f$ = Final angular velocity = $\dfrac{2}{1.33}\times 2\pi\ rad/s$

$\dfrac{I_f}{I_i}=\dfrac{1}{3}$

In this system the angular momentum is conserved

$L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s$

The angular velocity when the diver left the board is 3.14946 rad/s

3 0

3 years ago

On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86

Leto [7]

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

7 0

3 years ago