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love history [14]

3 years ago

15

A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30

m/s. What is the force exerted by the water jet

1 answer:

asambeis [7]3 years ago

6 0

Answer:

Explanation:

mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water

A = 3.14 x .005²

= 785 x 10⁻⁷ m²

mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg

momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .

Rate of change of momentum = 706.5 x 10⁻⁴

Let force be F

F - mg = 706.5 x 10⁻⁴

F = mg + 706.5 x 10⁻⁴

F = 23.55 x 10⁻⁴ x 9.8 + 706.5 x 10⁻⁴

= 937.3 x 10⁻⁴ N .

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A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis

Angelina_Jolie [31]

Answer:

1)

Force on bar magnet = 0

Torque on bar magnet = 0

2)

Force on bar magnet = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is inclined along z axis along magnetic field

then we will have

$\tau = MBsin0 = 0$

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is pointing 45 degree with z axis then we will have

$\tau = MBsin45$

$\tau = (2.5)(0.1)sin45$

$\tau = 0.177 Nm$

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is pointing 90 degree with z axis then we will have

$\tau = MBsin90$

$\tau = (2.5)(0.1)sin90$

$\tau = 0.25 Nm$

8 0

3 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago

Which of the following should NOT be discussed during safety training

Juli2301 [7.4K]

Is there suppose to be an image?

7 0

3 years ago

To determine if an object is in motion, you must use a __________. Question 1 options: Moving object Living thing Frame of Refer

Lubov Fominskaja [6]

Answer: moving object :)

Explanation:

8 0

3 years ago

Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay

Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

$w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s$

The maximum velocity is

$v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s$

8 0

3 years ago