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3 years ago

6

WHO WANTS BRAINLIEST THEN ANSWER THIS QUESTION

1 answer:

lozanna [386]3 years ago

6 0

if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...

thus the relative velocity will be 0

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What is the length x of the side of the triangle below? (Hint: use the cosine function.)

Firlakuza [10]

A because that is the answer

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2 years ago

Calculate the accleration of a car if its velocity increases from 15m/s to 75m/s in 5 second

andrezito [222]

Answer:

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3 years ago

A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee = 5kg

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Unknown:

Impulse of the frisbee = ?

Solution:

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3 years ago

Valence electrons determine an atom’s __________. a. mass c. atomic number b. chemical properties d. period Please select the be

Lemur [1.5K]

Hello,

Here is your answer:

The proper answer to this question is option B "<span>chemical properties".

Here is how:

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Your answer is B.

If you need anymore help feel free to ask me!

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5 0

2 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago