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3 years ago

50 POINTS!!! PLEASE!!!!!! Pls help me with this been stuck on it in like FOREVER!! PLSS.

Physics

2 answers:

Alenkinab [10]3 years ago

5 0

Answer:

it will move in 356.5g in diagonal direction(resultant or between the both direction)

Explanation:

285+428/2=356.5g

Hope u like it

PLZ MARK ME AS BRAINLIEST

Akimi4 [234]3 years ago

3 0

Backward. The most force is the 428 g block.

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A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr

astraxan [27]

Answer:

x = 0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

$Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2$

Now, we will calculate the force on the piston due to atmospheric pressure:

$Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N$

Now, for the compression of the spring we will use Hooke's Law as follows:

$F = kx\\$

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

8 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

3 years ago

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3. Which of the following statements about features on the sun is true?

Readme [11.4K]

Correct answer choice is:

b) Solar flares are spots hotter than the surrounding, therefore they are brighter than the rest, and they form where prominences start or end on the surface.

Solar flares eject coronal mass, they eject electrons, protons, and ions from the Sun. They<span> produce high energy particles and radiation that are dangerous to living organisms. </span>The earth encompasses a natural protection against these charged particles. Earth's magnetic field and atmosphere protect us from these particles.

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3 years ago

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A circuit can only light up a lightbulb if there is a ______ path for electricity to travel from one end of the energy source to

Dmitry_Shevchenko [17]

Answer:

Continuous

Explanation:

A circuit can only light up a lightbulb if there is a continuous path for electricity to travel from one end of the energy source to the other end.

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3 years ago

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A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

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