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Korvikt [17]
3 years ago
12

PLEASE HELP! I NEED ANSWER SOON

Physics
2 answers:
Serga [27]3 years ago
4 0

Answer:

5 and 3. I guess I am not very sure

Nikolay [14]3 years ago
4 0
The answers are 1,3,5
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During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
neonofarm [45]

Answer:

work is =7590joules

power = 23watts

4 0
3 years ago
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Two rods are made from materials that have different Young's moduli. The rods are constructed to have the same length and same c
ANTONII [103]

Answer:

This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.

Explanation:

σ=E*ε

ε=δ/L

σ=E*δ/L

δ=(σ*L)/E

σ=F/A

δ=(F*L)/(A*E)

As Force,Area and Length is same

δ∞1/E

From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.

5 0
3 years ago
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
Which of the following will cause a decrease in an object's weight? Increase in the mass of the object Decrease in the mass of t
Paul [167]
A decrease in mass will decrease an objects weight because 
weight = mass x gravitational constant
4 0
3 years ago
Read 2 more answers
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in
bezimeni [28]

1) Potential difference: 1 V

2) V_b-V_a = -1 V

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

\Delta U=q\Delta V

where

q is the charge's magnitude

\Delta V is the potential difference between the initial and final position

In this problem, we have:

q=4.80\cdot 10^{-19}Cis the magnitude of the charge

\Delta U = 4.80\cdot 10^{-19}J is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

V_b-V_a = - 1V

8 0
2 years ago
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