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ICE Princess25 [194]
2 years ago
11

A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100 N. The magnitude of the tension f

orce of each of the two supporting ropes is A : 190 N B : 120 N C : 60 N D : 94 N
Physics
1 answer:
guajiro [1.7K]2 years ago
3 0

Answer:

T = 190 N

Explanation:

When child is sitting on the swing then the weight of the child is vertically downwards

So it is

F_g = 160 N

now a force of 100 N is acting on the swing in horizontal direction

so it is given as

F_x = 100 N

now the net force is resultant force due to gravity and horizontal force

so it is given as

T = \sqrt{F_x^2 + F_g^2}

T = \sqrt{100^2 + 160^2}

T = 190 N

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A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was
Nookie1986 [14]

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

6 0
3 years ago
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
During which type of collision is none of the energy converted to heat
Gre4nikov [31]
This would be typical of an elastic collision.
4 0
3 years ago
Read 2 more answers
The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

delta(P)=F*delta(t)

P_{f} -P_{i}=F*delta(t)

2m(v_{f} -v_{i})=F*delta(t)

v_{i} =0.35-0.35=0

It is neccesary calculate the force:

F=(m+m)*g*sin\theta

Here, g = gravity = 9.8 m/s²

F=(0.2+0.2)*9.8*sin37=2.3591N

v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s

6 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
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