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2 years ago

8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?

1 answer:

pychu [463]2 years ago

5 0

(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²

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Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n

Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0

2 years ago

olga nikolaevna [1]

As he lifts the sack to his chest from the floor

5 0

2 years ago

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A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that

polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, $\theta=20^0$

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

$mg=2T\ sin\ \theta$

$T=\dfrac{mg}{2\ sin\theta}$

$T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}$

$T=936.7\ N$

Hence, this is the required solution.

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2 years ago

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What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?

kaheart [24]

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec )²× 0.20 m

F= 0.034 N

3 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

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2 years ago