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Hoochie [10]
3 years ago
9

Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the

crate if the mass of the crate is 20 kg?
5.5 m/s2
20 m/s2
17.5 m/s2
72 m/s2
Physics
1 answer:
nignag [31]3 years ago
5 0

Answer:

a = 5.5 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.

ΣF =m*a

where:

F = Force [N] (units of Newtons)

m = mass = 20 [kg]

a = acceleration [m/s²]

Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.

230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ]

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What conditions must be satisfied for a body to be in equilibrium <br>​
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For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero.

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3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What
Lubov Fominskaja [6]

Answer:

\frac{D}{d} = 4.12

Explanation:

As we know that resistance of one copper wire is given as

r = \rho \frac{L}{a}

here we know that

a = \pi (\frac{d}{2})^2

now we have

r = \rho \frac{L}{\pi (\frac{d^2}{4})}

r = \rho \frac{4L}{\pi d^2}

now we know that such 17 resistors are connected in parallel so we have

R = \frac{r}{17}

R = \rho \frac{4L}{17 \pi d^2}

Now if a single copper wire has same resistance then its diameter is D and it is given as

R = \rho \frac{4L}{\pi D^2}

now from above two equations we have

\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}

D^2 = 17 d^2

now we have

\frac{D}{d} = 4.12

3 0
3 years ago
A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea
Ray Of Light [21]

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² ×  sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

6 0
3 years ago
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

W = 25000/44

W = 568.18 N.

Hence the weight of the diver = 568.18 N

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3 years ago
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