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3 years ago

Brainliest for correct answer :)

Physics

2 answers:

AnnyKZ [126]3 years ago

8 0

All 3 objects have the same mass

AnnZ [28]3 years ago

4 0

Is the first . All 3 objects have the same mass

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A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago

How much kinetic energy will an electron gain if it accelerates through a potential difference of 23,000 volts in a cathode ray

Sauron [17]

Answer:

3.68x10-15 J

Explanation:

I hope this is the right answer sorry if im wrong

8 0

2 years ago

Please tell fast plzzzzzz

horrorfan [7]

Ovary the 2nd one is the answer

3 0

3 years ago

) The potential difference between the plates of a capacitor is 400 V. (a) If the spacing between the plates is doubled without

kramer

(a) 800 V

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 separation between the plates

We see that the capacitance is inversely proportional to the separation, d: in this problem, the separation between the plates is doubled (2d), so the capacitance will become half of its original value:

$C'=\frac{C}{2}$

The potential difference between the plates is related to the capacitance by

$V=\frac{Q}{C}$ (1)

where Q is the charge stored on the plate. In this problem, the charge is not changed: therefore, the new potential difference is

$V'=\frac{Q}{C/2}=2\frac{Q}{C}=2V$

So, the potential difference has doubled, and since the initial value was

V = 400 V

The new value is

V' = 2(400) = 800 V

(b) The charge will decrease by a factor 2

As before, the plate spacing is doubled, so according to the equation

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

Then the capacitance is halved again:

$C'=\frac{C}{2}$

This time, however, the voltage is held constant. We can rewrite the eq.(1) as

Q = CV

And since V has not changed, we can find what is the new charge stored in the capacitor:

$Q' = C'V=\frac{C}{2}V=\frac{1}{2}(CV) = \frac{1}{2}Q$

So, the charge will be halved compared to the original charge.

5 0

3 years ago

How many valence electrons do most stable atoms have?

IceJOKER [234]

Answer:

D) 8

Explanation:

Due to the octet rule the most stable atoms will have 8 valence electrons.

8 0

3 years ago