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wolverine [178]
3 years ago
15

Do the two cars ever have the same velocity at one instant of time? If so, between which two frames?Check all that apply. A. Car

s have the same velocity at one instant of time between dots 1 and 2. B. Cars have the same velocity at one instant of time between dots 2 and 3. C. Cars have the same velocity at one instant of time between dots 3 and 4. D. Cars have the same velocity at one instant of time between dots 4 and 5. E. Cars have the same velocity at one instant of time between dots 5 and 6. Cars never have the same velocity at one instant of time
Physics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

D. Cars have the same velocity at one instant of time between dots 4 and 5.

Explanation:

The two cars will eventually have the same velocity and it occurs between 4 and 5. Velocity is normally defined as displacement over time. This shows that a change in the value of either time or displacement will affect the value of the velocity. Thus, at a given time between the points 4 and 5, the two cars have the same value of velocity.

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If a light source (such as a light bulb) appears white, then it must _________ all colors of light.
pishuonlain [190]
Emit means put out so, emit
5 0
3 years ago
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
4 years ago
How much power does a hair dryer require to transform 198 000 J of energy in 15 mins?
Marina86 [1]

(198,000 joule / 15 minute) x (minute / 60 sec) = 220 joule/sec = <em>220 watts


</em>
7 0
3 years ago
Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
4 years ago
Try this out. Use a calculator!
Maslowich
Just subsitute and easy
v=55m/s
m=100kg
KE=(0.5)(100kg)(55m/s)^2
KE=(50kg)(3025 m^2/s^2)
KE=151250 J
2nd option
6 0
3 years ago
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