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3 years ago

An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster.

Physics

2 answers:

Dmitry [639]3 years ago

5 0

Answer:

640 J

Explanation:

ΔKE = ½ mv² − ½ mu²

ΔKE = ½ m (v² − u²)

ΔKE = ½ (80 kg) ((5 m/s)² − (3 m/s)²)

ΔKE = 640 J

Work = ΔE

Work = 640 J

Romashka [77]3 years ago

4 0

Answer:

640 for both

Explanation:

got right on edg

You might be interested in

Use the graph of velocity versus time for an object to answer the question.

Pavel [41]

The statement which fairly compares segment 2 and segment 3 is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Since segment 2 starts at t = 60 s and ends at t = 150 s, the time interval is Δt = 150 - 60 = 90 s.

Also, segment 3 starts at t = 150 s and ends at t = 240 s, the time interval is Δt = 240 - 150 = 90 s.

So, their time periods are the equal.

We notice that segment 2 is less steep than segment 3 this implies that the acceleration in each segment is different, since the acceleration is the slope of the graph.

Since force is determined by acceleration, this implies that the force on segment 2 is different form the force acting in segment 3.

So, we have equal time periods but different forces.

So, the statement which fairly compares segment 2 and segment 3 is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Learn more about velocity-time graph here:

brainly.com/question/24788847

8 0

3 years ago

47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

$N_1* x = N_2* (L - x)$

also we know that stress at both ends are same

$\frac{N_1}{12} = \frac{N_2}{8}$

$2*N_1 = 3*N_2$

Now from two equations we have

$\frac{3}{2}N_2*x = N_2* (L - x)$

solving above equation we have

$x = \frac{2}{5}L$

so the load is placed at distance 0.4L from the end of 12 mm^2 area

8 0

3 years ago

Question 1

Vesna [10]

Answer:

1) g = 4π² / m, 3) xaxis the length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

w = √ g / l

angular velocity, frequency and period are related

w = 2π f = 2π / T

we substitute

T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

T² = 4π² l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

y = (4π² / g) l

so the slope is

m = 4π² / g

clearing

g = 4π² / m

where the slope can be found with the values of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

m = 4pi2 / g

where it ends up reaching the floor

g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

5 0

4 years ago

G Railroad tracks are made from segments L = 79 m long at T = 20° C. When the tracks are laid, the engineers leave gaps of width

Andreyy89

Answer:

$l=L\alpha(T_c-T)$

Explanation:

L = Initial length of segment = 79 m

T = Normal temperature = $20^{\circ}\text{C}$

l = Width to be left for expansion

$\alpha$ = Coefficient of linear expansion of the material = $12\times 10^{-6}^{\circ}\text{C}^{-1}$

$T_c$ = Maximum temperature = $39.5^{\circ}\text{C}$

$\Delta T$ = Change in temperature = $T_c-T$

The expression of linear expansion is given by

$l=L\alpha\DeltaT\\\Rightarrow l=L\alpha(T_c-T)$

The expression for the minimum gap distance l the engineers must leave for a track rated at temperature $T_c$ is $l=L\alpha(T_c-T)$

8 0

3 years ago

Two wires are made from the same material. One wire has a resistance of 0.10 ω. The other wire is twice as long as the first wir

Anna71 [15]

Answer:

Explanation:

Given that,

First wire has a resistance of

R1 = 0.1 Ω

We are told that

Second wire is twice as long as the first wire.

Then,

Let the first wire has length

L1 = x

Then, second wire will have

L2 = 2x

Also, the radius of the second wire is half the radius of the first wire

Let the first wire has radius

r1 = y

Then, it area is

A1 = πr1² = πy²

Then, the second wire has radius

r2 = ½y

It area is also

A2 = πr2² = π(½y)² = ½πy²

Since the wire is made of the same material, then, they will have the same resistivity ρ

Then, we want to find the resistance of the second wire

Using

R = ρL/A

Where

R = resistance

ρ = sensitivity

L = length

A = area

Then, make resistivity subject of formula since it is a constant

RA = ρL

Then, ρ = RA/L

Then,

R1 • A1 / L1 = R2 • A2 / L2

Substituting each value

0.1 × πy² / x = R2 × ¼πy² / 2x

Cross multiply

0.1 × πy² × 2x = R2 ×¼π y² × x

Then,

R2 = 0.1 × πy² × 2x / ¼πy² × x

R2 = 0.1 × 2 / ¼

R2 = 0.8Ω

The resistance of the second wire is 0.8Ω

6 0

4 years ago