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3 years ago

A bicycle has a momentum of 23.4

Physics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Momentum=23.4kgm/s
Velocity=2m/s

$\\ \rm\longmapsto Momentum=Mass\times velocity$

$\\ \rm\longmapsto Mass=\dfrac{Momentum}{Velocity}$

$\\ \rm\longmapsto Mass=\dfrac{23.4}{2}$

$\\ \rm\longmapsto Mass=11.7kg$

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

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3 years ago

I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to

Vladimir79 [104]

Answer:

293k

Explanation:

In this question, we are asked to calculate the temperature to which the reaction must be heated to double the equilibrium constant.

To find this value, we will need to use the Van’t Hoff equation.

Please check attachment for complete solution

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3 years ago

A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h

Helen [10]

The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

Calculated value = 4.15 g/cm
accepted value = 8.64 g/cm

Substitute these values into equation 1

Error(%) = (4.15-8.64)100/8.64
Error(%) = -4.49(100)/8.64
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Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

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3 years ago

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Answer:

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3 years ago

Read 2 more answers

A lamp hangs from the ceiling at a height of 2.6 m. The lamp has a mass of 3.8 kg. The screws holding the lamp break, and it fal

iVinArrow [24]

Answer:

Explanation:

Given height of lamp from the ceiling = 2.6m

mass of the lamp = 3.8kg

acceleration due to gravity = 9.81m/s²

As the body falls to the ground, it falls under the influence of gravity.

Gravitational potential energy = mass*acc due to gravity * height

Gravitational potential energy = 3.8*2.6*9.81

Gravitational potential energy = 96.923 Joules

b) Kinetic energy = 1/2 mv²

m = mass of the body (in kg)

v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion $v^{2} = u^{2}+2gh$

$v^{2} = 0^{2}+2(9.81)(2.6) \\v^{2} = 51.012\\v =\sqrt{51.012}\\ v = 7.14m/s$

Since mass = 3.8kg

$K.E = 1/2 * 3.8 *7.14^{2}\\ K.E = 96.86Joules$

c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s

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3 years ago