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3 years ago

Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

Physics

1 answer:

choli [55]3 years ago

3 0

Answer:

$a=16\ m/s^2$

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

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A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

Where:

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In this problem we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

5 0

3 years ago

On the decibel scale,an increase of 1 dB means that sound intensity is increased by a factor of ________?

Maurinko [17]

Your answer would be 10 decibels

4 0

3 years ago

Read 2 more answers

Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at

Mashutka [201]

Explanation:

If Team A is on the left, B is on the right

if the force is constant, it means that the effort applied is equal.

So Team B is pulling 5000N to the right.

6 0

3 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$