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bixtya [17]
3 years ago
14

How much work is needed to lift a barbell that has a weight of 445 N a total of 2 meters off ground

Physics
1 answer:
AleksAgata [21]3 years ago
5 0

Answer:

Total work done to lift barbell = 890 J

Explanation:

Given:

Weight of barbell (F) = 445 N

Height (Distance) = 2 meter

Find:

Total work done to lift barbell = ?

Computation:

⇒ Work = Force(F) × Distance

⇒ Total work done to lift barbell = Weight of barbell × Distance

⇒ Total work done to lift barbell = 445 N × 2 meter

⇒ Total work done to lift barbell = 890 J

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. A mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N.
harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

7 0
3 years ago
Bob runs 1800 seconds at an average speed of 1.5 m/sec. How far did he go? 25 Points!!!!
german

Distance= Time×Speed

= 1800×1.5

= 2700 m

I am not sure it's right. the question itself is confusing.

4 0
2 years ago
5. What is the frequency of a sound wave when the speed of the sound is 340 m/s and has a wavelength of 1.21 m?​
LekaFEV [45]

Explanation:

v = f × /\

340 = 1.21f

f = 280.99Hz

3 0
3 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

5 0
3 years ago
8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o
Leni [432]

Answer:

The length of line is 78 cm or 0.78 m.

Explanation:

initial reading 2 mark

final reading 80 cm

The length of the line

= final reading - initial reading

= 80 - 2

= 78 cm

1 cm = 0.01  m

So, 78 cm = 0.78 m

4 0
2 years ago
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