A motor has a rating power of 20 W and motor is used for 10 s. Calculate the

1) Beth rode the Flying Saucer ride at her favorite amusement park. The ride hung from a rod and swung back and forth while simu

quester [9]

#1

Since Beth rode the Flying saucer ride which is spinning on its axis while axis is moving to and fro as well, so here Beth will experience two motions in this case

(i) Spinning motion

(ii) Rotational motion

#2

Jill and Scott both traveled for 30 min

So time of motion for both is 0.5 h as we know that 1 h = 60 min

Now from the formula of speed we know that

$speed = \frac{distance}{time}$

Speed of Jill

$v_1 = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v_2 = \frac{10}{0.5} = 20 km/h$

so correct answer is

B) Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

#3

Since we know that acceleration is rate of change in velocity

So here we can say that every motion where velocity changes with time then it is an accelerated motion.

Now change in velocity occurs when either the magnitude will change or its direction will change because here velocity is a vector quantity and it will change with magnitude and direction both.

so here due to oval path the direction of velocity will changes at many points and hence it will be an accelerated motion.

so correct answer is

C) James was right since the car changed direction during the race.

6 0

3 years ago

Read 2 more answers

An angle has a value of 2.2 radians. Find its value in degrees.

Irina18 [472]

Answer:

The value of 2.2 radians in degree is 126.11°.

Explanation:

Given that,

Angle = 2.2 radians

We need to calculate the value in degrees

We know that,

$1\ radians = \dfrac{180}{\pi}$

$2.2\ radians = \dfrac{180}{3.14}\times2.2$

$2.2\ radians = 126.11^{\circ}$

Hence, The value of 2.2 radians in degree is 126.11°.

8 0

3 years ago

Can light be reflected and refracted at the same time? if so, give an example

sp2606 [1]

sometimes you can see yourself as well, due to refraction and reflection

4 0

3 years ago

Read 2 more answers

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

3 years ago

A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i

Lubov Fominskaja [6]

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

$E_i = E_f$

$\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f$

initial height of the vaulter is equal to zero.

$\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f$

$h =\dfrac{v_i^2-v_f^2}{2g}$

$h =\dfrac{10^2-1^2}{2\times 9.8}$

h = 5.05 m

height of the jump is equal to 5.05 m.

4 0

4 years ago