a sports car can move 100 m in the first 4.4s of canstant accerationacceleration. what is the cars acceleration

vladimir1956 [14]

Answer:

a. petrification

Explanation:

tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)

amber = fossilized resin produced by extinct coniferous trees, typically yellow in color

mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying

3 0

3 years ago

Light traveling in water (n = 1.33) into an unknown medium.If rhe angles if incidence and refraction are 40 degrees and 25 degre

Bess [88]

40 degreased and 25 dregrees=65 degrees‍♀️

8 0

3 years ago

What happens to electrical energy that is used by objects in our homes? (1 point) a It is absorbed by batteries. b It is destroy

Reil [10]

Question:

What happens to electrical energy that is used by objects in our homes? (1 point)

a It is absorbed by batteries.

b It is destroyed.

c It is stored in solar panels.

d It is transformed into other forms of energy.

Answer:

D

3 0

3 years ago

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago

E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If

Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

$v_y = u_y +gt$

where:

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity (zero because the ball has been thrown horizontally)

$g=10 m/s^2$ is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

$v_y=0+(9.8)(0.6)=5.9 m/s$

b)

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

$v_x = 5 m/s$

8 0

4 years ago