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tigry1 [53]
3 years ago
13

A car accelerates from rest to 27 m/s in 8 seconds. What is the acceleration of the car?

Physics
1 answer:
Vlad1618 [11]3 years ago
3 0

Answer:

<h3>The answer is 3.38 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

where

a is the acceleration

v is the velocity

t is the time

From the question

v = 27 m/s

t = 8 s

We have

a =  \frac{27}{8}  \\  = 3.375

We have the final answer as

<h3>3.38 m/s²</h3>

Hope this helps you

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A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc
MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

distance=v\,*\, t

distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

6 0
3 years ago
Which of the following is not true about minerals?
mezya [45]

Answer:

I think it's A but I'm not sure

6 0
3 years ago
Read 2 more answers
A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

3 0
3 years ago
observe the figure given carefully volume of water in each vessel is shown arrange them in order of decreasing pressure at the b
mihalych1998 [28]

Answer:

See the explanation below

Explanation:

The pressure is defined as the product of the density of the liquid by the gravitational acceleration by the height, and can be easily calculated by means of the following equation.

 P=Ro*g*h

where:

Ro = density of the fluid [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation [m]

In this way we can understand that the greater pressure is achieved by means of the height of the liquid, that is, as long as the fluid has more height, greater pressure will be achieved at the bottom.

Therefore in order of decreasing will be  

The largest pressure with the largest height of the liquid, container B. The next is obtained with container D, the next with container A and the lowest pressure with container C.  

The pressure decreases as we go from the container B - D - A - C

5 0
3 years ago
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction \mu =0.4

Consider a small element at an angle \theta having an angle of d\theta

Normal Force=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}

N=T\cdot d\theta

Friction f=\mu \times Normal\ Reaction

f=\mu \cdot N

and T+dT-T=f=\mu Td\theta

dT=\mu Td\theta

\frac{dT}{T}=\mu d\theta

\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta

\frac{T_2}{T_1}=e^{\mu \pi}

\frac{T_2}{T_1}=e^{0.4\times \pi }

\frac{T_2}{T_1}==e^{1.256}

\frac{T_2}{T_1}=3.51

7 0
3 years ago
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