All categories

3 years ago

What is the tension in the cord after the system is released from rest? Both masses (A and B) are 10-kg.

Physics

1 answer:

lara31 [8.8K]3 years ago

4 0

Answer:

98 N.

Explanation:

Given data: mass= 10 kg, gravity= 9.8 m/s2

required: tension in the cord= ?

solution:

formula of tension= mass x gravity

by putting values of mass and gravity, we get

tension= 10 x 9.8

tension= 98 N. Ans

If the mass of the object which is attached with both ends of cord is 10 kg, so the tension which is a opposite force of weight is 98 N.

You might be interested in

A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t

OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

$(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s$

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

$s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s$

Now, the distance covered by the player in this time will be:

$s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)$

s₁ = 0.022 m

4 0

3 years ago

a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte

IgorLugansk [536]

From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.

dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

$1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.

7 0

3 years ago

The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric pote

tester [92]

Answer:

1.06 m

Explanation:

Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C

r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m = 1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m

The distance between them is 1.06 m

8 0

3 years ago

The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul

Ronch [10]

We Know, K.E. = 1/2 × m × v²
From the expression, we can conclude that Kinetic energy is directly proportional to mass. So, as mass will increase, Kinetic energy will also increase.

In short, Your Correct answer would be Option B

Hope this helps!

8 0

3 years ago

Read 2 more answers

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

Black_prince [1.1K]

Answer:

a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of the cylindrical shell

A = π r_b² - π r_a²

A = π (r_b² - r_a²)

we substitute

R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

V = i R

we subsist the expression of resistance

V = I ρ₀ L /π (r_b² - R_a²)

ρ₀ = V / I π (r_b² - R_a²) / L

6 0

3 years ago