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bonufazy [111]
3 years ago
10

Steven carefully places a m = 1.85 kg wooden block on a frictionless ramp so that the block begins to slide down the ramp from r

est. The ramp makes an angle of θ = 51.3 ° up from the horizontal. Which forces do nonzero work on the block as it slides down the ramp?
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

Gravity force

Explanation:

We are given that

Mass of wooden block,m=1.85 kg

\theta=51.3^{\circ}

We have to find the force which do non zero work on the block as it slides down the ramp.

When a block slides down the ramp on a frictionless surface .

Then, weight act on the wooden block in downward direction.

We know that weight is that force which act on the body due to gravity.

Then, we can say that the force which do nonzero work on the block as it slides down the ramp is gravity force.

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Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a
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The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

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We need to find the pressure exerted by camel feet.

As we know that :

{\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}

Substituting all the given values in the formula to find the pressure exerted by camel feet.

\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\  \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}}  \\  \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\  \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\  \\\star \:  \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}

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\rule{300}{2.5}

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2 years ago
You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x co
kvasek [131]

Answer:

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

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<u>(</u>2)---------- y - 61  = 0

y = 61.0 (3sf)

the new added vector is  = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

               = 22.5° (clockwise form -ve X axis)

<u />

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