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weqwewe [10]
2 years ago
10

A teacher performing demonstration finds that a piece of cork displaces 23.5 ml of water. The piece of cork has a mass 5.7 g. Wh

at is the density of the cork ?
Physics
1 answer:
Sedbober [7]2 years ago
8 0

Answer: 0.24g/ml

Explanation:

Given that:

Volume of water displaced = 23.5 ml

Mass of cork = 5.7 g

Density of the cork = ?

Recall that density is obtained by dividing the mass of a substance by the volume of water displaced.

i.e Density = Mass/volume

Density = 5.7g /23.5ml

Density = 0.24g/ml

Thus, the density of the piece of cork is 0.24g/ml

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Why is it unwise to stir a pot of soup with a metal spoon?
emmasim [6.3K]

Answer:

Explanation:

As we stir the pot of soup by a metal spoon which is a conductor, it conducts the heat and starts heating after some time the temperature of the spoon rises and we are not able to hold the spoon and we get burns.

3 0
2 years ago
Read 2 more answers
A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t
ivolga24 [154]

Answer:

1.32.225 N/C, direction is away from the point charge

2. 8.972*10^-12 C

3. the field is directed away from the axon

Explanation:

The electric field can be calculated as shown below:

E = k*|q|/r^2

Where:

E = electric field; k = 8.98755*10^9 N*m^2/C^2; r = distance between the measured field and point charge = 0.05 m; q = the point charge

For 0.100 m of the axon, the value of q is:

q = (5.6*10^11)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Thus:

q = (5.6*10^11)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Therefore:

E = (8.98755*10^9)*(8.972*10^-12)/0.05^2 = 32.255 N/C

A positive point charge always produce an electric field that is directed away from the field while a negative point charge produces an electric field that is directed toward the field

3 0
2 years ago
At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the
DENIUS [597]

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

7 0
2 years ago
The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu
Cerrena [4.2K]

Answer:

 maximumforce is F = mg

Explanation:

For this case we must use Newton's second law,

     Σ F = m a

bold indicate vectors, so we will write it in its components x and y

 X axis

       Fₓ = maₓ

 Axis y

      Fy - W = m aa_{y}

Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components

      Cos θ = Fₓ / F

      Fₓ = F cos θ

      sin θ = Fy / F

      Fy = F sin θ

Let's replace and calculate

      F sin θ -w = m a

 

As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)

       F sin θ = w = mg

The maximum value of this equation occurs when the sin=1, in this case

      F = mg

3 0
2 years ago
Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-
Nady [450]

Answer:

a)   W_total = 8240 J , b)  W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

       W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

       W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

          F - W = m a

          W = mg

          F = m (a + g)

          F = 80 (1 + 9.8)

          F = 864 N

The work of this force we will call it W1

   We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

        F₂ - W = 0

        F₂ = W

        F₂ = 80 9.8

        F₂ = 784 N

The work of this fura we will call them W2

The total work is

         W_total = W₁ + W₂

         W_total = (F + F₂) y

         W_total = (864 + 784) 5

         W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

        W₁ / W₂ = F y / F₂ y

        W₁ / W₂ = 864/784

        W₁ / W₂ = 1.1

7 0
2 years ago
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