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3 years ago

QUESTION 5

Physics

1 answer:

liberstina [14]3 years ago

5 0

Answer:

0.69 g/cm^3

Explanation:

Given that a small beaker has 50 mL of water in it. A small frog with a mass of 20 grams is dropped into the beaker. The water level rises to 79mL.

The volume of the frog will be:

79 - 50 = 29mL

Convert it to litre by dividing it by 1000

29/1000 = 0.029 L

Since 1L = 1000 cm^3

Convert it to cm^3 by multiplying it by 1000

0.029 × 1000 = 29 cm^3

Density = mass/volume

Substitute the mass and volume into the formula

Density = 20/29

Density = 0.6896 g/cm3

Therefore, the density of the frog is 0.69 g/cm^3 approximately

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The attractive or repulsive force between objects is called

Romashka [77]

The answer to this is electric force. Hope it helps

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3 years ago

Laura sees a horse pulling a buggy. She wonders how it can accelerate if the action of the horse pulling the cart would cause an

cricket20 [7]

Answer:

The net forces exerted on the horse and cart are not the same, so they are not balanced forces.

Explanation:

Using newton laws,

ΣF = ma

The two forces acting on the body is the forward force by the horse and the frictional force opposing the forward force

F-Fr= ma

Therefore,

F=ma+Fr

So the forward (by horse) is more than the opposite reaction (by cart).

The only time the forward force and the opposite reaction are the same is when a=0, I.e, the body is not moving at all or the body is moving at constant velocity.

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3 years ago

Read 2 more answers

A uniform solid sphere of mass M and radius R is free to rotate about a horizontal axis through its center. A string is wrapped

LenaWriter [7]

Answer:

$a = \frac{mg}{m + \frac{2}{5}M}$

Explanation:

To calculate the Acceleration and the tension of the object, we start by considering the value of the Tension through its moment of Inertia and Acceleration based on the angular velocity

$\tau = I\alpha = Tension(T)*R$

And $a = \alpha R$

Replacing,

$T*R = I\alpha = (\frac{2}{5} MR^2)*\frac{a}{R})\\T*R = \frac{2}{5}MaR\\T = \frac{2}{5}Ma$

The following forces occur in the body,

$mg - T = ma$

By this way we have the acceleration

$mg - \frac{2}{5}Ma = ma$

$a(m + \frac{2}{5})M) = mg$

$a = \frac{mg}{m + \frac{2}{5}M}$

6 0

4 years ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

BabaBlast [244]

Answer:

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.
friction between the skater and the ice is negligible.

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = $\frac{1}{3} mL^{2}$

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I = $\frac{1}{3}$ x 4.81 x $0.7^{2}$ = 0.79 kg-m for each arm

2. Her body as a cylinder has moment of inertia = $\frac{1}{2} mr^{2}$

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = $\frac{1}{2}$ x 64.38 x $0.175^{2}$ = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk

I = (0.79 x 2) + 0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = $\frac{2\pi }{60}$ x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m

angular momentum = Iω

= 0.99 x ω = 0.91ω

according to conservation of angular momentum, both angular momentum must be equal, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷ $\frac{2\pi }{60}$ = 176.38 rpm

7 0

3 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago