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3 years ago

7

A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the

ground?

2 answers:

Sedaia [141]3 years ago

8 0

....................................................................
64 meters

Ivenika [448]3 years ago

4 0

64 meters from the base of the cliff.

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A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

Charra [1.4K]

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$

$W_y = 500 sin37$

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

$F = W_y = 500 sin37$

$F = 300 N$

so it apply 300 N force along the inclined plane

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Understanding that if we say something unkind to someone else his or her

OlgaM077 [116]

Answer: Moral

Explanation:

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The value of 1.0004 to the power 1 by 2 using Binomial approximation is

IgorLugansk [536]

Given:

The given value is $(1.0004)^{\frac{1}{2}}$ .

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

$(1.0004)^{\frac{1}{2}}$

It can be written as:

$(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}$

$(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004$ $[\because (1+x)^n=1+nx]$

$(1.0004)^{\frac{1}{2}}=1+0.0002$

$(1.0004)^{\frac{1}{2}}=1.0002$

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An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice

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Answer:

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Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

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