How many electrons would it take to make up 1 C of charge? ( Coulomb’s law )

A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the

viva [34]

Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

2as = u²

s = u² / 2a

s = (14)² / (2 x 16.333)

s = 6.0 m

Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.

The correct option is d

4 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a

zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

$\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0

3 years ago

Read 2 more answers

The least common fossils are those that have been<br> petrified<br> frozen<br> buried<br> distilled

Alchen [17]

Answer: Frozen fossils

6 0

4 years ago

If you jog at a speed of 1.5m/s for 20 seconds how far di you travel

Harlamova29_29 [7]

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

Must click thanks and mark brainliest

3 0

3 years ago