A 200kg bucket of cement

Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the

Marrrta [24]

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised from T to T'.

Hence specific heat of a substance is calculated as-

$c= \frac{Q}{m[T'-T]}$

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

7 0

3 years ago

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The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

3 years ago

Can you explain what a mixture is from a chemical prospective?

sergeinik [125]

Answer:

Mixtures are materials that contain two or more chemical substances dispersed among each other (mixed together). If no chemical reaction occurs when two materials are mixed, they form a mixture. The chemical properties of the components don't change.

Explanation:

5 0

2 years ago

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Which side of the electromagnetic spectrum has shorter wavelengths?

PIT_PIT [208]

Answer:

Hey,

All the answers should be blue.

Edg 2020/2021

8 0

3 years ago

A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr

PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
Now, we need to ask ourselves: what supplies this force?
In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
We know that this force can be expressed as follows:

$f_{frs} = \mu_{s} * F_{n} (1)$

where μs = coefficient of static friction between the rock and the merry-

go-round surface = 0.7, and Fn = normal force.

In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
Fn = m*g (2)
This static friction force is just the same as the centripetal force.
The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

$F_{c} = m* \omega^{2}*r (3)$

Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

$\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec$

This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.

4 0

3 years ago

A 200kg bucket of cement​

A 200kg bucket of cement