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2 years ago

9

A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th

e acceleration

2 answers:

I am Lyosha [343]2 years ago

8 0

Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>

OverLord2011 [107]2 years ago

6 0

Answer:

Acceleration, $a=6\ m/s^2$

Explanation:

Given that,

Initial speed of the car, u = 35 m/s

Final speed of the car, v = 65 m/s

Time taken, t = 5 s

We need to find the acceleration of the car. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{65-35}{5}$

$a=6\ m/s^2$

So, the acceleration of the car is $6\ m/s^2$ . Hence, this is the required solution.

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You hold a piece of wood in one hand and a piece of iron in the other. both pieces have the same volume, and you hold them fully

aivan3 [116]

You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully under water at the same depth. At the moment you let go of them, which one experiences the greater buoyancy force?<span>
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2 years ago

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0

2 years ago

1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

4. A cylindrical tube has a length of 14.4cm and a radius of 1.5cm and is filled with a colorless gas. If the density of the gas

professor190 [17]

Answer:

Mass, m of gas is 0.2504 grams.

Explanation:

First, we need to solve for the volume of the cylindrical tube.

Volume of cylinder is given by the formula;

$V = 2\Pi r^{2}h$

Where, V represents volume.

π represents pie

r represents radius.

h represents height or length.

Given the following data;

Radius, r = 1.5cm

Length, h = 14.4cm

Density, d = 0.00123g/cm³

Substituting into the equation;

$V = 2 * 3.142 * (1.5)^{2}14.4$

$V = 2 * 3.142 * 2.25 * 14.4$

$V = 203. 6016$

Therefore, the volume of the cylindrical tube is 203. 6016cm³

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

$Density = \frac{mass}{volume}$

$Mass = density * volume$

Substituting into the equation, we have;

$Mass = 0.00123 * 203. 6016$

Mass = 0.2504g.

5 0

2 years ago