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3 years ago

Describe 3 ways a bicyclist can change velocity.

2 answers:

8 0

She can pedal harder to increase the speed of the bicycle. She can apply the brakes to decrease the speed of the bicycle. And she can turn the handle bars to change the bicycle's Direction.

ivann1987 [24]3 years ago

4 0

1. Applying more force and energy, speeding up
2. Releasing potential and kinetic energy by going down a slope, speeding up
3. Applying less force to slow down and let friction overcome the bicyclist

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11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75

lora16 [44]

Answer:

Explanation:

Find the diagram relating to the question for proper explanation of the question below.

Using the principle of moment

Sum of clockwise moments = Sum of anticlockwise moments

Moment = Force * perpendicular distance

For anti-clockwise moment:

Since the 30 kg moves in the anticlockwise direction according to the diagram

ACW moment = 30 * 1 = 30 kgm

For clockwise moment

If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).

Equating both moments we have;

0.75M = 30

M = 30/0.75

M = 40 kg

The second child's mass is 40 kg

5 0

3 years ago

A rocket is dropped out of an airplane at 100 m/s (downward). If the rocket fires causing an upward acceleration of Ct2 and it t

Mkey [24]

Answer:

C = 0.0125 m/s⁴. The calculation procedure can be found in the attachment below. The concept of motion along a straight line with constant acceleration has been applied to solve the problem.

Explanation:

The sign convention chosen in this problem solution is upwards as positive and downwards negative. The equation of motion v = u + at has been used to calculate the constant C as only one unknown is contained in this equation. This is so because we have been given the initial velocity, the acceleration and the time taken. To solve future problems of this kind, first thing to check for is an equation of motion with the least number of unknown. This helps to reduce the complexity of the problem solution.

5 0

3 years ago

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef

bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0

3 years ago

What’s the answer?hhhu

allsm [11]

Which one if you want independent variable meaning that means variable (often denoted by x ) whose variation does not depend on that of another.

7 0

3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago