Ask question

Ask question

All categories

2 years ago

15

A bag of sugar weights 20 N on the earths surface. If you double the distance from the center of the earth, the bag now weighs w

hat

1 answer:

elena-s [515]2 years ago

4 0

I have no idea that’s a hard one

You might be interested in

Order the speed of sound through these materials from the slowest to the fastest.

Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0

2 years ago

An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

2 years ago

Using the picture above, which ball has the greatest potential energy?

weqwewe [10]

Ball 4 because the higher the elevation is the greater the potential energy it has

4 0

2 years ago

The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a

Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

$J=F.\Delta t$

$J=10^6\ N\times 20\ s$

$J=2\times 10^7\ N-s$

(b) Impulse is also given by the change in momentum i.e.

$J=\Delta p=p_f-p_i$

$J=p_f-p_i$

$p_f=J+p_i$

$p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s$

$p_f=20143000\ kg-m/s$

(c) For new velocity,

$v_f=\dfrac{p_f}{m}$

$v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}$

$v_f=1549.46\ m/s$

Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

so

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago