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Luda [366]
3 years ago
12

Why do you have to take mass into consideration when doing the egg drop

Physics
1 answer:
PSYCHO15rus [73]3 years ago
6 0

Answer:

the higher the mass the faster it falls so you need to know the mass to know how fast or slow it will fall

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The odometer of a car changes from 1048 km to 1096 km in 40
Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0
2 years ago
Calculate the acceleration of a bottle rocket that is traveling at a speed of 74 m/s, and then descends at a speed of 89 m/s. It
Harlamova29_29 [7]

Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.

Explanation:

5 0
3 years ago
Water has a specific heat capacity nearly nine times that of iron. Suppose a 50-g pellet of iron at a temperature of 200∘C is dr
miv72 [106K]

Answer:

a. closer to 20∘C

Explanation:

m_{p} = mass of pallet = 50 g = 0.050 kg

c_{p} = specific heat of pallet = specific heat of iron

T_{pi} = Initial temperature of pellet = 200 C

m_{w} = mass of water = 50 g = 0.050 kg

c_{w} = specific heat of water

T_{wi} = Initial temperature of water = 20 C

T_{e} = Final equilibrium temperature

Also given that

c_{w} = 9 c_{p}

Using conservation of energy

Energy gained by water = Energy lost by pellet

m_{w} c_{w} (T_{e} - T_{wi}) = m_{p} c_{p} (T_{pi} - T_{e})\\(0.050) (9) c_{p} (T_{e} - 20) = (0.050) c_{p} (200 - T_{e})\\ (9) (T_{e} - 20) =  (200 - T_{e})\\T_{e} = 38 C

hence the correct choice is

a. closer to 20∘C

4 0
4 years ago
two soccer players are running for the same ball during a game. Without looking the two collide. One player has a mass of 45 kg
jeka57 [31]

Well yuh see I dont got a clue

6 0
4 years ago
Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold
arsen [322]

Answer:

Resistance of gold wire, R=1977 \times 10^3 ohm

Explanation:

In this question we have given

Density of gold, d=19.3\times 10^3 \frac{kg}{m^3}

resistivity of gold, r=2.44\times 10^{-8} ohm.m

Length of wire, L= 2.05 km

Temperature, T= 20^oC

We know that relation between volume and density is given as

Density= \frac{mass}{Volume}

Therefore, volume occupied by one gram gold is given as,

V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3.........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

V=\pi \times r^2 \times L......(2)

Here,

A= \pi \times r^2...........(3)

here A is the cross sectional area of cylendrical gold wire  

From equation 2 and 3

we got

V=A \times L...............(4)

on comparing equation 1 and equation 4, we got,

A \times L=5.181\times 10^{-8} m^3

A=\frac{5.181\times 10^{-8} m^3}{2050 m}

A=2.53\times 10^{-11}m^2

we know that resistance and resistivity are related by following formula,

Resistance = resistivity\times \frac{L}{A}................(5)

Put values of resistivity, A and L in equation 5, we got

R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}

R=1977 \times 10^3 ohm

Therefore resistance of gold wire, R=1977 \times 10^3 ohm

7 0
3 years ago
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