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3 years ago

How much work does it take to accelerate a 9.0 kg object from rest to 27 m/s?

1 answer:

5 0

Initial kinetic energy = 0J (v=0 (rest))

Final kinetic energy = 1/2mv ²
Ek=1/2(9)(27 ²)
Ek=4.5(729)
Ek=3280.5

∆Energy=3280.5J
(As it starts from 0)

Work= ∆energy
So work =3280.5J

Answer=3300J (forth option)

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In a butcher shop, a horizontal steel bar of mass 4.94 kg and length 1.46 m is supported by two vertical wires attached to its e

mojhsa [17]

Answer:

Tension in right wire = 25.9N

Explanation:

I have attached a free body diagram to depict this question.

From the diagram, i have labelled the tensions in the strings T1 and T2.

While i labelled the weight of the bar as Wb and weight of sausage as Ws.

Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)

Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.

Thus;

T1 + T2 = Wb + Ws

We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N

Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N

Thus,

T1 + T2 = 48.46N + 24.43

T1 + T2 = 72.89N - - - - - (eq 1)

Now, let's take moments about the left end of the bar.

The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m

So, moments about left end;

T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)

1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)

1.46T2 = 35.373 + 2.443

1.46T2 = 37.816

T2 = 37.816/1.46 = 25.9N

3 0

3 years ago

Weight is influenced by which of the following

nignag [31]

Out of the given options, weight is influenced by mass and gravity

Answer: Option A

Explanation:

The object's mass is defined as the quantity of a matter with which the object is formed. It can change its state of matter but the quantity will remain the same. However, the weight is defined as how much force gravity exerts on the object's mass to pull it.

The mass is always same irrespective the location but the weight may vary from one place to the other while talking for the bigger picture. For example, the object's weight may be 60 kg on Earth but when it is measured on the moon, it will be lesser.

The weight of an object generally has nothing doing with the volume and it doesn't depend solely on the gravitational pull. The mass plays a crucial role.

$W=F=m \times g$

6 0

3 years ago

1-Calculate Req

ArbitrLikvidat [17]

Answer:

1. 21.66 Ohms

2. 3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps. I am not an expert in physics but its ok :)

Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.

3 0

2 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

Which wavelength of light is capable of penetrating the dust of a nebula?

ivann1987 [24]

When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.

8 0

3 years ago