All categories

2 years ago

First, record some data for this comparison in the table below.

Physics

1 answer:

kondor19780726 [428]2 years ago

5 0

Answer:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

Explanation:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

This is the answer

You might be interested in

A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(

maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0

2 years ago

Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors commit

Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

= $\dfrac{difference}{original}\times 100$

= $\dfrac{63-42}{63}\times 100$

= $\dfrac{21}{63}\times 100$

= 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0

2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0

2 years ago

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0

2 years ago

Read 2 more answers

The standing vertical jump is a good test of an athlete's strength and fitness. The athlete goes into a deep crouch, then extend

SashulF [63]

Answer:

<em>The athlete will rise 1.10 meters off the ground</em>

Explanation:

<u>Vertical Motion</u>

If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by

$\displaystyle y_m=\frac{v_o^2}{2g}$

The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at

$\displaystyle a=\frac{F_n}{m}=\frac{2W}{m}=2g$

The speed at the end of his push can be computed by

$v^2=2ay$

Replacing the value of a obtained above:

$v^2=4gy$

where y is the length of this crouch

$v^2=4\cdot 9.8\cdot 0.55$

$v=4.64\ m/s$

This is the initial speed of this vertical launch, thus

$\displaystyle y_m=\frac{4.64^2}{2\cdot 9.8}$

$y_m=1.10\ m$

5 0

2 years ago