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zavuch27 [327]
3 years ago
10

Austin performed an experiment. He put 100mL of vegetable oil (density 0.9 g/ml) in a graduated cylinder. He put a 100g mass of

an unknown metal in to the oil and it sank while displacing 11 mL of oil. How much should the unknown metal weigh in the oil?
Physics
1 answer:
Grace [21]3 years ago
6 0

Answer:

Explanation:

mass of displaced oil = 11 x  .9

= 9.9 gm

9.9 x 10⁻³ kg

weight of displaced oil = 9.9 x 9.81 x 10⁻³ N

= .097 N .

buoyant force by oil = .097 N

weight of unknown metal = .1 x 9.8

= .98 N .

weight of metal in oil = .98 - .097

= .883 N .

=

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How is the surface area of an average person is 2m^2 in the chapter pressure
11Alexandr11 [23.1K]

Answer:

We have to show surface area =2m^2,with few conditions that is by considering Force =200000\ N and Pressure =100000\ Pa to be respectively.

Explanation:

The atmospheric pressure is =10^{5}\ Pa on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is =2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton.

Now to calculate surface area we can find it from pressure=\frac{force}{area} and re-arranging it to.

area=\frac{force}{pressure}

So plugging the values,

Surface area =\frac{20000}{10000}=2\ m^{2}

Hence from the above calculations we can say that surface area is 2m^2.

So the surface area of an average person can be said to have 2m^2, using the concept of pressure and force.

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2 years ago
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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0
2 years ago
Two forces are going in opposite directions each force is 9.
professor190 [17]

Answer:

The net force is zero.

Explanation:

Two opposing and equal forces cancel each other out, giving you a net force of zero.

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3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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