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LenKa [72]
2 years ago
11

A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m

/s^2.the body will reach the surface if earth in how much time.
Physics
1 answer:
butalik [34]2 years ago
7 0
First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s
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Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

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3 0
2 years ago
1. An object is projected upward with a velocity of 125 m/s.
grigory [225]

Answer:

A) s = 796.38 m

B) t = 12.742 s

C) T = 25.484 s

Explanation:

A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.

v = u + gt

u = 125 m/s

v = 0 m/s

g = 9.81 m/s²

Thus;

0 = 125 - 9.81(t)

g is negative because motion is against gravity. Thus;

9.81t = 125

t = 125/9.81

t = 12.742 s

Max height will be gotten from Newton's 2nd equation of motion;

s = ut + ½gt²

s = (125 × 12.742) + (½ × -9.81 × 12.742²)

s = 1592.75 - 796.37

s = 796.38 m

B) time to reach maximum height is;

t = u/g

t = 125/9.81

t = 12.742 s

C) Total time elapsed is;

T = 2u/g

T = 2 × 125/9.81

T = 25.484 s

4 0
3 years ago
Needddd helppppppp!!!
yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0
2 years ago
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