1. An object is traveling at a constant velocity of 30 m/s when it experiences a constant

A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

7 0

3 years ago

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop

hoa [83]

Refer to the attachment

5 0

3 years ago

Which of the following are true statements? A. Like charges repel B. Unlike charges repel C. Unlike charges attract or D. Charge

Brums [2.3K]

Answer: The correct answers are (A) and (C).

Explanation:

The expression from electrostatic force is as follows;

$F=\frac{kq_{1}q_{1}}{r^{2} }$

Here, F is the electrostatic force, k is constant, r is the distance between the charges and $q_{1},q_{1}$ are the charges.

The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.

Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.

The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.

Therefore, the true statements from the given options are as follows;

Like charges repel.

Unlike charges attract.

7 0

3 years ago

What is an analogy of two different roads or rivers to compare a series and parallel circuit?

faltersainse [42]

Answer:

In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.

5 0

2 years ago

Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the fi

Delicious77 [7]

Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the time (T) it takes for the first drop to strike the floor

from s = ut + $0.5at^{2}$

200 = 0 + $0.5 x 9.8 x T^{2}$

200 = $4.9 x T^{2}$

200 / 4.9 = $T^{2}$

T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3}T)^{2}$ )

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2}$ )

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{1}{3}T)^{2}$ )

s = 0 + ( $0.5 x 9.8 x (\frac{1}{3} x 6.4)^{2}$ )

s = 22.22 cm

6 0

3 years ago