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Veseljchak [2.6K]

3 years ago

12

A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.

1 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops

1 answer:

shutvik [7]3 years ago

5 0

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

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Two objects have the same momentum. Which of the following is true? I. The masses of the objects are equal. Ii. The speeds of th

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Answer:

Iv. The direction of motion of each object is the same.

Explanation:

Momentum of an object is defined as:

$p= mv$

where

m is the mass of the object

v is its velocity

Since v is a vector, it follows that p is also a vector, and it has the same direction of the velocity. This means that if two objects have same momentum, then the direction of their momentum vector is also the same: therefore, the direction of motion of each object must be the same.

The other options are wrong, because they just state that only one of the two quantities involved in the equation (mass or velocity) is the same, but the two objects can actually have the same momentum even if they have different masses and velocities (in fact, the only thing that matters is that the product between mass and velocity is the same).

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3 years ago

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Answer:

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2 years ago

A snowball that will be used to build a snowman is at the top of a only hill. If the

Darina [25.2K]

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

$E=mgh$

g is acceleration due to gravity

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$h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m$

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3 years ago

A 50kg girl is playing soccer. She runs towards the ball at a speed of 7 m/s. How much kinetic energy does she have as she runs

stepladder [879]

The calculation for kinetic energy is this

KE = 1/2mv^2

KE = 1/2(50)(7^2)

KE = 1/2(49•50)

KE = 1225 kgm^2/s^2.

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3 years ago

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ( $Q$ ), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ( $U_{l,ice}$ , $U_{l,steam}$ ), in joules, and two sensible component ( $U_{s,w}$ ), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

$c_{ice}$ - Specific heat of ice, in joules per kilogram per degree Celsius.

$c_{w}$ - Specific heat of water, in joules per kilogram per degree Celsius.

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$ , $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$ , $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$ , $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$ , then the amount of thermal energy is:

$Q = 15167500\,J$

The amount of thermal energy needed is 15167500 joules.

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