The horizontal distance travelled by the package is 180.8 m.
What is the time of motion of the package?
The time of motion of the package is the time taken for the package to travel from its initial position to the given height.
The time taken for the package to travel through the given distance is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity of the package = 0t is time of motionh is the height through which the package travelledg is acceleration due to gravityh = ¹/₂gt²
2h = gt²
t = 2h/g
t = √(2h/g)
t = 4.52 seconds
The horizontal distance travelled by the package is calculated as follows;
X = Vxt
where;
Vx is the horizontal velocity of the packaget is the time of motionX = (40 m/s) x (4.52 s)
X = 180.8 m
Thus, the distance the package strike relative to where it was released is the horizontal displacement of the package.
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Two students are on a balcony 19.6 m above the street. One student throws a ball vertically down ward at 14.7 m / s At the same instant, the other stu dent throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down . a. What is the difference in the time the balls spend in the air? b . What is the velocity of each ball as it strikes the ground ? c . How far apart are the balls 0.800 s after they are thrown ?
According to the given statement:
a)The difference in the time the balls spend in the air is 3s.
b)Each ball travels at a speed of 24.5 m/s as it lands.
c) The balls 0.800 s after they are thrown far apart are 23.52 m.
What exactly is meant by velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In principle, velocity is a vector quantity. It is the velocity where at distance changes. It is the migration change rate.
Briefing:A )
We have u = 14.7 m/s, acceleration = acceleration due to gravity = 9.8 m/s² , we need to find time when s = 19.6 m.
The first ball:
h = v o × t - g t² / 2
19.6 = 14.7 t + 9.8 t²/2
4.9 t² + 14.7 t - 19.6 = 0
t 1/2 = (-14.7 +/- √216.09+384.16 ) / 9.8
= 9.8/9.8 = 1 s
The second ball:
v = v o - g t
0 = 14.7 - 9.8 t
9.8 t = 14.7
t = 1.5 ( upward )
h = 19.6 + 14.7 - 9.8 × 1.5 = 30.625 m
30.625 = 9.8 × t²/2
t² = 6.25
t = 2.5 s ( downward )
1.5 s + 2.5 s = 4 s
The difference: 4 s - 1 s = 3 s.
B )
We have u = 14.7 m/s, acceleration = acceleration due to gravity = 9.8 m/s² , and t = 1 second.
The first ball:
v = v o + g t
= 14.7 + 9.8 × 1
= 24.5 m/s
The second ball:
v = g * t = 9.8 * 3 = 24.5 m/s
Each ball travels at a speed of 24.5 m/s as it lands.
C ) d 1 = 14.7 * 0.8 + 9.8 * 0.8²/2
= 11.76 + 3.136
= 14.896 m
d 2 = 14.7 * 0.8 - 9.8* 0.8²/2
= 11.76 - 3.136
= 8.624 m
d 1 + d 2 = 14.896 + 8.624 = 23.52 m
The balls 0.800 s after they are thrown far apart are 23.52 m.
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A passenger on a boat moving at 2.50 m/s on a still lake walks up a flight of stairs at a speed of 0.60 m/s (Figure 1). The stairs are angled at 45∘ pointing in the direction of motion as shown. What is the magnitude of the velocity of the passenger relative to the water?
The magnitude of the velocity of the passenger on a boat moving at 2.50 m/s on a still lake relative to the water is 2.95 m / s
Velocity of man = 0.6 m / s
Velocity of ship = 2.5 m / s
Resolving the velocity of man walking at an angle of 45° into its X and Y components.
X-components of velocity of man = 0.6 * cos 45°
X-components of velocity of man = 0.42 m / s
Y-components of velocity of man = 0.6 * sin 45°
Y-components of velocity of man = 0.42 m / s
[tex]V_{x}[/tex] = 0.42 + 2.5
[tex]V_{x}[/tex] = 2.92 m / s
[tex]V_{y}[/tex] = 0.42 m / s
V² = [tex]V_{x}[/tex]² + [tex]V_{y}[/tex]²
V² = 2.92² + 0.42²
V² = 8.53 + 0.18
V = √ 8.71
V = 2.95 m / s
Therefore, the magnitude of the velocity of the passenger relative to the water is 2.95 m / s
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The volume of asamplea saof nitrogen gas is 88mLat 25°C and 1.0 atm. Whatis its volume in LatSTP?V₁=88mL T₁=25°CV₂=? T₂=23°2
Given:
V1= 88 ml
P1= 1 atm
T1= 25°C= 298.15 K
STP conditions:
P2= 1atm
T2= 273.15 K
Apply :
[tex]\frac{P1V1}{T1}=\frac{P2V2}{T2}[/tex]Replacing:
[tex]\frac{1(88)}{298.15}=\frac{1(V2)}{273.15}[/tex][tex]88(273.15)=298.15V2[/tex][tex]\frac{88(273.15)}{298.15}=V2[/tex]V2= 80.62 ml
Suppose you apply a flame and heat one liter of water, raising its temperature 10 degrees celsius. If you transfer the same heat energy to two liters, how much will the temperature rise?
Transferring the same heat energy to two litres the temperature will rise by 5 degree celsius.
What is temperature and how the temperature rise to be 5 degree celsius?Temperature is the measure of degree of hotness and coldness , to either measure the temperature of room or weather.Here in this question we are applying a flame and heat one litre of water by which the temperature is raised 10 degree celsius.Now we are transferring the same heat energy to two litres , then the temperature rise is asked.The d , density of water =1000 kg/m^3 , using the equation Q= mCdelT , where C = 4.18 KJ.The heat we got is 41.8 KJ, the change is temperature comes out to be 5 degree celsius.To know more about temperature visit:
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A vertical wire, suspended from one end is stretched by attaching a weight of 20N to the lower end. If the weight extends the wire by Imon. Calculate the energy gained
A vertical wire suspended from one stop is tensioned by attaching a 20N weight to the lower stop. If you have weight, use Imon to stretch the wire. Energy source: `W=(1)/(2) f xx l=(1)/(2) xx 20 xx 10^(-3)` `= 0.01J`.
In physics, power is a quantitative property that is transferred to an identifiable frame or physical device in the form of labor productivity and warmth and tenderness. Electricity is a stored quantity. Energy conservation regulations state that energy can be transformed into form, but cannot be created or destroyed.
Energy exists in various characteristic documents. Examples include light energy, thermal energy, mechanical energy, gravity, electrical energy, sound electricity, chemical power, and nuclear or nuclear energy. Each form can be transformed or transformed into a different bureaucracy. Power is in everything we consume, eat, and use. Power supplies and regulates the body's natural internal abilities. It is used to support cells and tissues in the body, build muscle, and is important for maintaining homeostasis. The harsher the environment, the more energy it takes to maintain.
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