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ki77a [65]
3 years ago
13

Which of the following are non-contact forces: friction, electrostatic force, magnetic force, gravity?

Physics
2 answers:
oee [108]3 years ago
4 0
Magnetic force and Gravity are non-contact forces among friction, electrostatic force, magnetic force and gravity.
CaHeK987 [17]3 years ago
4 0
Gravity and Magnetic force are non-contact forces 
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P waves can travel through Earth's outer core, but S waves cannot. Because of this, scientists know
eduard

Answer:

The answer is B

please mark me brainliest

Explanation:

5 0
2 years ago
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A stone is thrown vertically straight up with a velocity of 2800cm/s . By neglecting the air resistance , find the maximum heigh
yarga [219]
We know that t=2.8571428571428568
and therefore 28*2.8571428571428568 - 9.8*2.8571428571428568^2/2
is around 40 m
6 0
3 years ago
The intensity of a certain siren is 0.060 W/m^2 from a distance of 25 meters. What is the intensity of this siren at a distance
Nady [450]
<h2>The intensity of siren is 0.015 W m⁻²</h2>

Explanation:

The intensity of sound is inversely proportional to the square of distance between source and the observer .

Thus Intensity at a distance 25 m  is I₁ ∝ \frac{1}{r_1^2}                               I

Similarly the intensity at a distance 50 m is I₂ ∝ \frac{1}{r_2^2}               II

Dividing II by I , we have

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{25^2}{50^2} = \frac{1}{4}

Thus I₂ = \frac{1}{4} x 0.06 =  0.015 W m⁻²

7 0
3 years ago
A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal
ss7ja [257]

Answer:

v_{o}=141.51m/s

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

x=1420m\\g=-9.8m/s^{2}

From the equation of x-position we know that

x=v_{ox}t=v_{o}cos(35)t

Solving for v_{o}

v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)} (1)

Now, if we analyze the equation of y-position we got

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0

0=v_{o}sin(35)t+\frac{1}{2}gt^{2}

Knowing the equation for v_{o} in (1)

0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}

\frac{1}{2}(9.8)t^{2}=1420tan(35)

Solving for t

t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s

Now, we can solve (1)

v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s

6 0
3 years ago
Read 2 more answers
If an object has zero acceleration, does it have to have zero velocity?
adelina 88 [10]

Answer:

Yes, the velocity would also be zero.

Explanation:

Acceleration is the change in velocity over time, therefore, there has to be a change in velocity for something to accelerate. which means without acceleration, the object has no velocity.

3 0
3 years ago
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