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3 years ago

Will mark as brainliest !! plz answer it

Physics

2 answers:

MrRissso [65]3 years ago

5 0

Don’t listen to the other guy. I answered on your other comment!

Alchen [17]3 years ago

4 0

Answer:

1/4

Explanation:

pls mark me as brainliest thanks

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Which of the following may have happened if the inner planets had greater masses during the formation of the solar system? . . A

Svetach [21]

Right answer is option b that is the inner planet may havebeen larger and more gaseous.

5 0

3 years ago

The moon has a mass of 7.35x1022 kg and is a lot farther away than is shown in textbooks. The mass of the

anyanavicka [17]

Answer:

$F=1.98\times 10^{20}\ N$

Explanation:

Given that,

The mass of a Moon, $M_m=7.35\times 10^{22}\ kg$

The mass of the Earth, $M_e=5.98\times 10^{24}\ kg$

The moon's mean orbit distance around the earth is, $r=3.84\times 10^8\ m$

We need to find the gravitational force exerted on the moon by the Earth.

The formula of gravitational force is given by :

$F=G\dfrac{M_mM_e}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.35\times 10^{22}\times 5.98\times 10^{24}}{(3.84\times 10^8)^2}\\\\F=1.98\times 10^{20}\ N$

So, the required force is $1.98\times 10^{20}\ N$ .

3 0

3 years ago

How much water remains unfrozen after 62.2 kJ is transferred as heat from 364 g of liquid water initially at its freezing point?

Amanda [17]

Answer:

177.213 grams

Explanation:

Given:

Total amount of water = 364 grams

Latent heat of fusion = 333 kJ/kg

Heat transferred = 62.2 kJ

The amount of water frozen = 62.2/333 = 0.186786 kg = 186.786 grams

Hence, the amount of water remained unfrozen = Total water - Amount of water frozen

the amount of water remained unfrozen = 364 grams - 186.786 grams = 177.213 grams

6 0

4 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

When physical activity is a set plan or program for all ages, usually performed by a group through a local recreational unit, it

Alexeev081 [22]

Answer:

C. Sports activity

Explanation:

A sport activity is a form of competitive physical activity in forms of games. They are normally organized for individuals to participate with the aim of improving physical abilities and skills, bring enjoyment to participants and act as a form of entertainment.

8 0

3 years ago