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algol [13]
3 years ago
6

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

Physics
2 answers:
mario62 [17]3 years ago
7 0
A :-) work = force x distance
W = 85 x 15
W = 1275 joules

.:. The work done is 1275 joules
denpristay [2]3 years ago
4 0

Answer:

1275J

Explanation:

Given parameters:

Force on box  = 85N

Distance moved  = 15m

Unknown:

Work done  = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

 Work done  = Force x distance

Now insert the parameters and solve;

 Work done = 85 x 15  = 1275J

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If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property
Ainat [17]

Answer:

  • <u><em>soluble</em></u>

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a<em> physical property</em> because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is <em>soluble</em> is describing a <em>physical property.</em>

3 0
3 years ago
5.<br>why does the pointer rotate?​
morpeh [17]

Answer: find the answer in the explanation.

Explanation:

From the experiment set up in the diagram, the pointer is resting on the drinking straw while the rod is resting on the drinking straw.

When the rod is being heated through the bursen burner, there will be linear expansion in the rod. As the rod increases its length, this causes the drinking straw to roll and thereby causing the pointer to rotate.

The pointer therefore rotates because of the thermer expansion that happen in the rod due to the heat from the bursen burner.

8 0
3 years ago
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?
lina2011 [118]
We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
\sum F = ma

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

mg \sin \theta - \mu_K N = ma (parallel direction) (1)
mg \cos \theta - N =0 (perpendicular direction) (2)
where
\theta=35^{\circ} is the angle of the inclined plane, N is the normal reaction of the plane, \mu_K N is the frictional force, with \mu_K=0.4 being the coefficient of friction.

From eq.(2), we find
N=mg \cos \theta
and if we substitute into eq.(1), we can find the acceleration of the block:
mg \sin \theta - \mu_k mg \cos \theta = ma
from which
a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2
7 0
3 years ago
A neuron has a resting potential of about _____ millivolts.
Leno4ka [110]
A neuron has a resting potential of about 70mV (millivolts)
5 0
3 years ago
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