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Vladimir [108]
3 years ago
13

Starting from rest, a car takes 2.4s to travel 15m. Assuming a constant acceleration, how long will it take the car to travel th

e next 15m?
Physics
1 answer:
Alborosie3 years ago
7 0

Answer:

T = 2.4 + 2.4 = 4.8 [s]

Explanation:

In order to solve this problem, we must use the following kinematics equation and calculate the acceleration value.

x=x_{o} +v_{o}*t+(\frac{1}{2})*a*t^{2}

Vo = inital velocity = 0

x - xo = 15 [m]

t = time = 2.4 [s]

15 = 0.5*a*(2.4)^2

a = 5.208 [m/s^2]

We can use the same equation to find the time.

30 = 15 + 0.5*(5.208)*t^2

t = 2.4 [s]

T = 2.4 + 2.4 = 4.8 [s]

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What is the difference between an elastic and inelastic collision
barxatty [35]

Answer:

In an elastic collision, the total kinetic energy is conserved, while in an inelastic collision, it is not

Explanation:

Let's define the two types of collision:

- Elastic collision: an elastic collision is a collision in which:

1) the total momentum of the system is conserved

2) the total kinetic energy of the system is conserved

Typically, elastic collisions occur when there are no frictional forces acting on the objects in the system, so that no kinetic energy is lost into thermal energy. An example of elastic collision is the collision between biliard balls.

- Inelastic collision: an inelastic collision is a collision in which:

1 ) the total momentum of the system is conserved

2) the total kinetic energy of the system is NOT conserved

In an elastic collision, part of the total kinetic energy is lost (=converted into thermal energy) due to the presence of frictional forces. An example of inelastic collision is the accident between two cars, in which part of the energy is converted into heat.

4 0
4 years ago
is the science that manages the database, digital tools, and software used by both medicine and forensic science to store and an
Ne4ueva [31]

Explanation :

Bioinformatics is the science that manages the database, digital tools and software used by both medicine and forensic science to store and analyze DNA.

It involves the use of computers to collect all the data and organize data. It also develops methods and computational tools for understanding biological data.

It helps in tracing the evolution of organism using DNA and builds a computational model.

8 0
3 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another
Harman [31]

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

4 0
2 years ago
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
Gekata [30.6K]

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

7 0
3 years ago
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