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Valentin [98]
2 years ago
11

Please help ASAP- A book has a mass of 5 kg, a baby has a mass of 5 kg. Which one weighs more?

Physics
2 answers:
Sphinxa [80]2 years ago
7 0

Answer:

Same mass

Explanation:

5kg and 5kg are the same that's the explanation

mihalych1998 [28]2 years ago
4 0
They weigh the same.
Weight = mass * gravity
W of the baby = 5 * 9.81
W of the book = 5 * 9.81
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"One step in the manufacture of silicon wafers used in the microelectronics industry is the melt crystallization of silicon into
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7 0
3 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

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7 0
2 years ago
Use your knowledge of waves to explain why echoes occur. Use your explanation to devise a system to measure distances to objects
german

Explanation:

Echoes occur due to the reflection of sound from any obstacle, but not all the reflected sound waves lead to the phenomenon of echo. For the echo to be heard it actually depends upon the human perception as well, human ears can encounter the difference between the sound wave directly form the source and the reflected sound waves only if there is a minimum time gap of one-tenth of a second. For this time gap in the atmosphere at normal temperature and pressure the obstacle must be at least 7 meters away from the sound source.

3 0
3 years ago
An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
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