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Zielflug [23.3K]
3 years ago
10

During a test crash an airbag inflates to stop a dummy's forward motion. the dummy's mass is 75kg. If the net force on the dummy

is 825n toward the rear of the car , what is the dummy's deceleration.
Physics
1 answer:
Black_prince [1.1K]3 years ago
4 0
F=ma
a=F/m
=825N/75kg
=825kg*m/75kg*s^2
=11m/s^2 in the direction of the force (ans)
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A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

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3 years ago
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zhannawk [14.2K]
Average speed= total distance/total time =12km/h
6 0
3 years ago
Describe the principle of conversation of energy:
exis [7]

Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another

Explanation:

The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.

Practical examples include:

- electrical pressing iron that converts electrical energy to heat energy

- solar panels that converts solar energy to electrical energy

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7 0
3 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

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