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storchak [24]
3 years ago
6

If a car is moving on a highway at 70 km/hr, going South and then turns to travel 70 km/hr East. What happens to the cars speed

and velocity?
(A)Speed - changing, Velocity - constant


(B)Speed - constant, Velocity - changing


(C)Speed - constant, Velocity - constant


(D)Speed - changing, Velocity - changing
Physics
1 answer:
frozen [14]3 years ago
6 0

Answer:

(B)Speed - constant, Velocity - changing

Explanation:

The magnitude of the velocity ( speed ) has not changed as car is traveling constantly at 70 km/h. But we know the velocity is also having a direction component, and the car which was traveling South is now traveling East. As direction is changed, velocity is changed but as magnitude of velocity has not changed, the speed remain unchanged.

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Roughly speaking, the radius of an atom is about 10,000 times greater than that of its nucleus. If an atom were magnified so tha
skad [1K]

Answer:

0.124 miles

Explanation:

Since the radius of the atom R = 10000r where r = radius of nucleus. Now, if r = 2.0 cm = 0.02 m,

R = 10000r

= 10000 × 0.02 m

= 200 m

We know that 1 mile = 1609 m.

So the radius of the atom in miles is R = 200 m × 1 mile/ 1609 m = 0.124 miles

So, the radius of the atom in miles is 0.124 miles  

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What is cell in electricity?
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Answer:

An electrical cell is a device used to generate electricity, or to make chemical reactions by applying electricity.

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3 years ago
A 0.350kg bead slides on a curved fritionless wire,
LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

m1*g*h-\frac{m1*V_B^2}{2}=0  Solving for Vb:

V_B=\sqrt{2gh}=6.56658m/s

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

m1*g*h2-\frac{m1*V_{B'}^2}{2}=0 Solving for h2:

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6 0
3 years ago
a baseball player is dashing toward home plate with a speed of 200 feet per hour. When he decides to hit the dirt. he slides for
Serggg [28]

To solve this problem we will use the kinematic formula for the final velocity.


V_f_x = V_0_x + a_xt


The final speed is 0 at the moment the player stops.


The time until it stops is 1.3 s


The initial speed is 200 feet / s  Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.


a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.



To answer part b) we use the following formula.


Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet

8 0
3 years ago
the maximum displacement of a particle, within a wave, above or below its equilibrium position, is called__________
Gre4nikov [31]

Answer:

the amplitude.

Explanation:

3 0
3 years ago
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