Answer:
x = 129.9 m
y = 30.9 m
Explanation:
When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.
Given data:
= 50 m/s
Angle = 30°
Time = t = 3 s
horizontal component of velocity = 
= cos30°
= 50cos30°
= 43.3 m/s
Vertical component of velocity = 
= Sin30°
= 50Sin30°
= 25 m/s
This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.
But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.
Horizontal distance = x = t
x = 43.3*3
x = 129.9 m
Vertical Distance = y = t -0.5gt²
y = 25*3 - 0.5*9.8*3²
y = 75 - 44.1
y = 30.9 m
Answer:
0.53 quart
Explanation:
The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C
Since, ΔV = VγΔθ
substituting the values of the variables into the equation, we have
ΔV = VγΔθ
ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C
ΔV = 528900 × 10⁻⁶ quart
ΔV = 0.528900 quart
ΔV ≅ 0.53 quart
Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.
<span>Waves cause erosion through the processes of impact and abrasion. </span><span>
2. c. they move up and down, but do not move forward. </span>
The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
<h3 /><h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
Pressure is found as the product of the density,acceleraton due to gravity and the height.
P₁=ρ₁gh₁
P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m
P₁=24014.88 N/m²
P₂=ρ₂gh₂
P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m
P₂=196.2 N/m²
P₃=ρ₃gh₃
P₃=850 kg/m³×9.81 (m/s²)×0.25
P₃=2084.625 N/m²
Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
To learn more about the pressure refer to the link;
brainly.com/question/356585
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