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2 years ago

If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?

Physics

1 answer:

kicyunya [14]2 years ago

6 0

For this case we have that by definition:

$F = ma$

Where,

m: mass of the object
a: acceleration of the object
F: summation of forces

We have then:

$F = 10 + 5\\F = 15 N$

Then, by clearing the acceleration we have:

$a = \frac {F} {m}$

Substituting values we have:

$a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}$

Answer:

The acceleration of the box is equal to:

$a = 3 \frac {m} {s ^ 2}$

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An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The cur

MariettaO [177]

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$

is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

$\Delta T=100-21\ ^0C=79\ ^0C$

So,

$\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ$

Heat Supplied $Q=VIt$

where $i=current$

$V=Voltage$

$8.5\times 120\times t=2.4\times 4186\times 79$

$t=778.10 s$

6 0

2 years ago

A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

adelina 88 [10]

14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.

6 0

2 years ago

Read 2 more answers

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

2 years ago

What does the term equilibrium refer to?

snow_lady [41]

<h3>Answer;</h3>

A) the resting position of the wave

<h3>Explanation;</h3>

A wave is a transmission of a disturbance from one point which is the source to another, and this involves transfer of energy through a material medium.
Equilibrium refers to a state of balance between opposing forces, it is a state of balance in which opposing forces cancel one another. 
When wave is in rest position its called equilibrium position of a wave. When a wave travels through a material medium, the particles in the medium are disturbed from their resting, or equilibrium positions.

5 0

2 years ago

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

mV + MV = mu + MU

V(m+M) = mu

V = mu/(m+M)

Substituting the values in the above equation,

V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)

= 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0

2 years ago