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4 years ago

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A sled with no initial velocity accelerates at rate of 5.0 m/s^2 down a hill . How long does it take the sled to go 45 m to the

bottom

1 answer:

Leya [2.2K]4 years ago

6 0

The time taken by the sled to reach 45 m to the bottom is 4.24 s.

The sled's initial velocity <em>u</em> is zero. It has an acceleration <em>a</em> of a value 5 m/s² down the hill and it travels a distance <em>s</em> equal to 45 m down the hill.

Use the equation of motion,

$s=ut+\frac{1}{2} at^2$

Substitute 0 m/s for u and rewrite the equation for t.

$t=\sqrt{\frac{2s}{a}} \\ =\sqrt{\frac{2(45 m)}{5.0m/s^2} } \\ =4.24s$

The sled takes a time of 4.24 s to reach the bottom of the hill

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A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of char

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Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux, $\phi=-2 EA=-2E(\pi r^2)$

$\phi=-2\times 300\times (3.14\times (0.66)^2)$

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c.

Net charge, $Q=\epsilon_0\times \phi$

Where

$\epsilon_0=8.85\times 10^{-12}$

$Q=-820.67\times 8.85\times 10^{-12}$

$Q=-7.26\times 10^{-9} C$

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3 years ago

How do you calculate force?

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Force (N) = mass (kg) × acceleration (m/s²)

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4 years ago

In Hooke's law, Fspring=kΔx , what does the Fspring stand for?

Mrrafil [7]

Answer:

The amount of force acting on the spring.

Explanation:

Here, Fspring= kΔx is a equation denoting the amount of force acting on the spring.

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3 years ago

Read 2 more answers

A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

finlep [7]

Answer:

The acceleration expressed in the new units is $114.048 Km/h^{2}$

Explanation:

To convert from $m/s^{2}$ to $Km/h^{2}$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}$

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

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4 years ago

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;

$\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)$

Therefore;

$\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)$

From equation (4) we now make the new force $F_{o2}$ the subject of formula as follows;

${F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}$

$Q_o$ then cancels out from both side of the equation, hence we obtain the following;

$3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)$

From equation (4) we can now write the following;

$F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}$

This could also be expressed as follows;

$F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2$

3 0

4 years ago