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Dafna11 [192]
2 years ago
9

Example 2.1

Physics
1 answer:
tatiyna2 years ago
8 0

Answer:

5 min 20 km --------

Explanation:

5 min 20 km --------

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Byeeeeeeeeeeeejdiiwidjwdjj2xi2idiwidisididie
Advocard [28]

Answer:

Be serious or else CEO of brainly will kick you out

3 0
2 years ago
The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f
Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0
3 years ago
A magnetic pickup is used to detect the shaft speed of an experimental high speed compressor. a gear with 12 teeth is used to ex
worty [1.4K]
The answer would be 39 seconds
5 0
3 years ago
Read 2 more answers
A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t
Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, \epsilon=1.5\ V

Magnitude of Earth's field, B=10^{-4}\ T

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}

\omega=159.15\ rad/s

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0
3 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
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