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kirza4
3 days ago
9

Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t

he mass and v is the speed (velocity). Assume that the speed of the soda bottle falling from a height of 0.8 m will be 4 m/s, and use this speed for each calculation. Record your calculations in Table A of your Student Guide. When the mass of the bottle is 0.125 kg, the KE is kg m2/s2. When the mass of the bottle is 0.250 kg, the KE is kg m2/s2. When the mass of the bottle is 0.375 kg, the KE is kg m2/s2. When the mass of the bottle is 0.500 kg, the KE is kg m2/s2.
Physics
2 answers:
kicyunya [1K]3 days ago
9 0

Answer:

The values are 1, 2, 3, and 4 corresponding to the edge.

Explanation:

ValentinkaMS [1.1K]3 days ago
6 0

Kinetic energy is represented as

KE = (0.5) m v²

In each scenario, v = the velocity of the bottle set at  4 m/s

with m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

for m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

if m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

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Pilobolus is a genus of fungi commonly found on dung and known for launching its spores a large distance for a sporangiophore on
Softa [913]

1) The work performed on the spores is 2.45\cdot 10^{-7}J

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy principle, the work done on the spores equals their change in kinetic energy:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

W is work done

m represents the spore's mass

v is the final velocity

u is the initial velocity

Given:

spore mass = m=10^{-8} kg

initial velocity u = 0 (starting from rest)

final velocity v = 7.0 m/s

Calculating the work:

W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J

2)

The spores follow projectile motion, moving along a parabolic trajectory made of two motions:

- Constant speed horizontally

- Vertically accelerated due to gravity

Considering vertical motion first, using kinematics:

s=ut+\frac{1}{2}at^2

where

s = 0.01 m (shooting height)

u = 0 (no initial vertical velocity, horizontal ejection)

t = time of flight

g = acceleration due to gravity a=g=10 m/s^2

Solving for t:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.01)}{10}}=0.045 s

Now for horizontal displacement, when velocity is constant:

d=v_x t

where

horizontal velocity = v_x = 7.0 m/s

time t = 0.045 s

Calculating distance d:

d=(7.0)(0.045)=0.32 m

Thus, the spores land 0.32 meters away.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

Learn more about projectile motion:

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3 0
13 days ago
A cart moves along a track at a velocity of 3.5 cm/s. when a force is applied to the cart, its velocity increases to 8.2 cm/s. i
inna [987]
In physics, acceleration describes how an object's velocity changes over time. Whenever the speed of an object shifts, it's considered to be accelerating. To calculate acceleration here:

acceleration = (8.2 - 3.5) / 1.5 = 3.1 m/s²

I trust this answers your question.
5 0
14 days ago
Read 2 more answers
Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements
Yuliya22 [1153]

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

8 0
12 days ago
A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
inna [987]
Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

3 0
6 days ago
-. What is the acceleration of 4 kg trolling bag pulled by a girl with a<br> force of 3 N?
inna [987]

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

3 0
14 days ago
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