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Keith_Richards [23]
2 years ago
14

Can someone please help me with this one

Physics
1 answer:
Brilliant_brown [7]2 years ago
5 0

Answer:

in left

Explanation:

Hope it will help

<em>p</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>m</em><em>a</em><em>r</em><em>k</em><em> </em><em>a</em><em>s</em><em> </em><em>a</em><em> </em><em>b</em><em>r</em><em>a</em><em>i</em><em>n</em><em>l</em><em>i</em><em>s</em><em>t</em><em>s</em>

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Is a scientist who studies climate change. As part of a larger experiment, her work requires her to travel. Her company makes su
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3 years ago
A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-&gt; of magnitude 4.8
n200080 [17]

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

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3 years ago
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soldi70 [24.7K]
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7 0
3 years ago
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

3 0
3 years ago
The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,
Tamiku [17]

Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: I=80dB and r_{1}=2m

we need an intensity level of: I_{2}=40dB

thus, we are looking for the distance r_{2}.

which we can find with the law for intensity and distance:

(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}

we solve for r_{2}:

\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }

and we substitute the known values:

r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m

at a distance of 2.83m the intensity level is 40dB

5 0
3 years ago
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