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andrew11 [14]
3 years ago
14

A car travels in a straight line for 4.4 h at a constant speed of 81 km/h. What is its acceleration?

Physics
2 answers:
Fed [463]3 years ago
7 0

The speed and length of time don't matter.

Constant speed in a straight line is the DEFINITION of ZERO acceleration.

irga5000 [103]3 years ago
5 0

Answer:

a = 0m/s^2

Explanation:

we can use the equation:

v_f = a(t) + v_i

And plug in our numbers:

81km/h = am/s^2(4.4h) + 81hm/h

0 = am/s^2(4.4h)

0 = am/s^2

a = 0m/s^2

Because the speed was constant, there was no difference between the final and initial velocities, meaning that once you do the math you see there is no acceleration affecting the speed.

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A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)
egoroff_w [7]

Explanation:

The given data is as follows.

             q = 6.0 nC = 6 \times 10^{-9} C

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            \sigma = \frac{q_{in}}{A} .......... (1)

Since, area of the sphere is as follows.

               A = 4 \pi r^{2} ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      \sigma = \frac{q_{in}}{4 \pi r^{2}}

                   = \frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}            

                   = 0.477 \times 10^{-5}

or,               = 4.77 \mu C/m^{2}

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 \mu C/m^{2}.

5 0
3 years ago
A 2-column table with 5 rows. The first column titled metal has entries aluminum, cork, iron, lead, wax. The second column title
astraxan [27]

Answer:Cork and wax

Explanation:

9 0
3 years ago
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A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is
Blizzard [7]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

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In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

##############################

If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

_________________________________

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5 0
2 years ago
If an experiment involves a large volume of liquid a _______ would most likely be used to hold it.
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The answer is d a beacker
3 0
2 years ago
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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

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3 years ago
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