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ra1l [238]
2 years ago
14

A roller coaster is moving at 25m/s at the bottom of a hill. Three seconds later it reaches the top of the hill moving at 10m/s.

What was the acceleration of the coaster
Physics
1 answer:
marishachu [46]2 years ago
8 0

Answer:

5 m/s^{2}

Explanation:

acceleration is defined as the rate of change of velocity per unit time. Therefore, acceleration, a=\frac {\triangle v}{\triangle t}

where \triangle v is change in velocity and t is time

Substituting initial and final velocities with 25 m/s and 10 m/s then using time as given in the question of 3 s then

a=\frac {25-10}{3}=\frac {15}{3}=5 m/s^{2}

Therefore, the acceleration is 5 m/s^{2}

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If you push on the floor with 5 N for 15 seconds, how much work is done?
Maslowich

Answer: It would be 75

Explanation: If your pushing on the floor at a rate of 5 for 15 seconds you would multiply the numbers and get 75

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3 years ago
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Hola tengo un taller de física y nose como resolver este pregunta
Blizzard [7]

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

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

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

8 0
2 years ago
I don't understand this question please help
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I think it is coal.  I also know the US has lots of natural gas as well.
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2 years ago
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Is Particle B charged or uncharged? If charged, indicate whether it is charged positively or negatively. (n = neutron, p = proto
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Answer:

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Explanation:

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7 0
3 years ago
What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g
aalyn [17]

Answer:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle \theta, given by:

f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)

Where:

Z_1 e represent the charge of the projectile (Z1=2)

Z_2 e is the target charge (z2=79)

K= 5x10^6 eV represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

t= 10^{-8] m represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

n =\frac{\rho N_A N_M}{M_g}

Where:

\rho = 19.3 g/m^3= 19300 Kg/m^3

N_A = 6.022 x10^{23} molecules/mol the Avogadro's number

N_M = 1 represent the atoms per molecule

M_g = 197 g/mol = 0.197 Kg/mol represent the molecular weigth

If we replace we got:

n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness t=10^{-8}m

And using the first formula we got:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

4 0
3 years ago
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