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Alex
3 years ago
12

Ent Cards 4.Q: True or False? Anotherword for tube is "duct." A: Explanation:

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0
No it would be pipe or metro
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For a complete vector quantity, what information needs to be given?
zhenek [66]

The magnitude of the quantity and its direction.

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2 years ago
Will water pressure be greater at the bottom of a container that you put a hole in or at the top of a container?
ddd [48]
Pressure=hrg
pressure is great at the bottom because the weight lf water exerts pressure below the container due to the gravitational pull of the earth.
4 0
3 years ago
Power is work done over a what?
melomori [17]
No the answer is energy 
4 0
3 years ago
A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel
lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: -10 [N] = m.ax

Y-direction balance force: -m*9,8 \frac{m}{s^2} = m.ay

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

ax=(-10 [N]) / m

ay=-m*9,8\frac{m}{s^2} / m

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

ay=-9,8\frac{m}{s^2}

For the x component aceleration we must replace the Newton unit:

N =\frac{kg.m}{s^2}

ax= -10 \frac{kg.m}{s^2} / (2 kg)

ax= - 5 \frac{m}{s^2}

6 0
3 years ago
An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
zzz [600]

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

8 0
3 years ago
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