If a bicycle that is traveling at 5 mph speeds up to 10 mph in 5 seconds, the bicycle's acceleration will be 0.45m/s²
HOW TO CALCULATE ACCELERATION:
The acceleration of the bicycle can be calculated by dividing the change in velocity by time. That is;a = (v - u) ÷ t
Where;
a = acceleration
v = final velocity (m/s)
u = initial velocity (m/s)
t = time (s)
According to this question, a bicycle that is traveling at 5 mph speeds up to 10 mph in 5 seconds10mph = 4.47m/s5mph = 2.24m/sa = (4.47 - 2.24) ÷ 5
a = 2.23 ÷ 5
a = 0.45m/s²
Therefore, a bicycle that is traveling at 5 mph speeds up to 10 mph in 5 seconds, the bicycle's acceleration will be 0.45m/s²Learn more at: https://brainly.com/question/24696936?referrer=searchResults
Assume that you are in a space station conducting fluid mechanics experiments in zero gravity. You perfectly coat a sphere and a long cylinder with a thin layer of the same fluid and thickness. If the sphere and cylinder have the same radius, will the fluid layers on the two objects have the same pressure
Answer:
Explanation:
From the Question operation is
-zero gravity
-same fluid thickness
-same radius
Generally the force is given by [tex]f=ma=\frac{p}{a}[/tex]
=> [tex]g=m/s^2=0[/tex]
=> As acceleration varies directly to pressure
=>[tex]p=0[/tex]
Generally it is shown that the fluid layer on the two objects will have same pressure.
One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is at rest as it hangs from the string. When the object is raised to a position X that is a height H above its lowest point and released from rest, the object undergoes simple harmonic motion. When the object passes through the equilibrium position Y, it has a speed v0.
What methods could a student use to determine the total mechanical energy E at position Y, and why?
Answer:
v₀ = [tex]\sqrt{2gH}[/tex]
to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
Explanation:
In this internal exercise the student must use the conservation of mechanical energy,
Starting point. Highest point of the trajectory
Em₀ = U = m g H
Point of interest. Point at height Y
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g Y
energy is conserved
Em₀ = Em_{f}
m g H = ½ m v² + m g Y
v² = 2 g (H -Y) (1)
in this case they indicate that Y is the equilibrium position whereby Y = 0 and the velocity is v = v₀
v₀ = [tex]\sqrt{2gH}[/tex]
Therefore, to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
charges that are different each other
Answer:
Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.
Answer:
positive and negative ??
Explanation:
I dont really understand the question if im being honest
a glass flask of volume 400 cm cube is just filled with Mercury at zero degree Celsius how much marks are you will overflowing the temperature of system Rises up to 80 degrees celsius
Answer:
5.76 cm³
Explanation:
Using the equation for volume expansivity,
V = V₀ + V₀γΔθ) where V₀ = volume of cube = volume of mercury = 400 cm³, γ = cubic expansivity of mercury = 18 × 10⁻⁵ /K and Δθ temperature change = θ₂ - θ₁ where θ₁ = 0 °C and θ₂ = 80°C. So, Δθ = 80°C - 0°C = 80°C = 80 K
Now, the volume change, ΔV = V - V₀ = V₀γΔθ.
So, substituting the values of the variables into the equation, we have
ΔV = V₀γΔθ
= 400 cm³ × 18 × 10⁻⁵ /K × 80 K
= 5.76 cm³
So the mercury will overflow by 5.76 cm³.
Consider two atomic clocks, one at the GPS ground control station near Colorado Springs (elevation 1830 m) and the other one in orbit in a GPS satellite (altitude 20200 km). According to the general theory of relativity, which atomic clock runs slower?
a. The orbiting clock alternately runs slow and then fast depending on where the Sun is.
b. The clock in orbit runs slow. The clock in Colorado runs slow.
c. The orbiting clock is 95% slower than the clock in Colorado.
d. The clocks keep identical time.
Answer:
b. The clock in orbit runs slow. The clock in Colorado runs slow.
Explanation:
b. The clock in orbit runs slow. The clock in Colorado runs slow. This is because of the fact that as the gravity weakens, time moves faster, hence, the answer is b.
Answer: The clock in Colorado runs slow
Explanation: I got it correct on the HW. It should be a separate answer from B though.
Looking at the bank statement, what three items total that same amount?
The balance sheet, income statement, and cash flow statement each offer unique details with information that is all interconnected. Together the three statements give a comprehensive portrayal of the company's operating activities.
Here's a quick story if a boy walks to the woods at night and slender man comes along what is most likely to happen put your answers in the comments.
Answer:
the boy would go missing
Answer: Slender man converts the boy to christianity
what is the best inference for the speed of the car after 3 seconds
Answer:
B. 17m/s
Explanation:
This question contains a graph that illustrates the relationship between the speed of a car over time. The graph shows that one can make an inference of the amount of time it takes for the car to cover a particular speed and vice versa.
In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.
Below are 3 pieces of cobalt metal . In each, four different magnetic domains are drawn. Each domain is labeled with an arrow indicating the direction of the magnetic field in that domain. Which of these bars of cobalt metal has a magnetic field?
O Bar A
O Bar B
O Bar C
While looking at bromine (Br) on the periodic table, a student needs to find another element with very similar chemical properties but with a different atomic mass. Which best describes the student’s options?
There are two possible elements, and they are directly to the left and right of bromine.
There are two possible elements, and they are directly above and below bromine.
There are many possible elements, and they are all in the same vertical column as bromine.
There are many possible elements, and they are all in the same horizontal row as bromine.
Answer: 3) There are many possible elements, and they are all in the same vertical column as bromine.
Explanation:
Answer:
choice 3
Explanation:
A 74.0-kg man stands on a bathroom scale inside an elevator.(a) The elevator accelerates upward from rest at a rate of 1.10 m/s2 for t1 seconds. What does the scale read during this time interval
Answer:
806.6 N
Explanation:
Given that :
Mass (m) = 74 kg
Acceleration due to gravity (g) = 9.8 m/s²
Acceleration during interval t1 = 1.10 m/s²
Using the relation :
Normal reaction = Mass (g + acceleration at interval t1)
Normal reaction = 74(9.8 + 1.10)
Normal reaction = 74(10.9)
Normal reaction = 806.6 N
An object moves uniformly around a circular path of radius 19.0 cm, making one complete revolution every 2.40 s.
a. What is the translational speed of the object?
b. What is the frequency of motion inhertz.
c. What is the angular speed of the object?
Answer:
v = 0.5 m/s
f = 0.42 Hz
ω = 2.6 rad/sec
Explanation:
By definition, the translational speed is the rate of change of the position with respect to time.The change in position along a complete revolution is just the following:Δs = 2*π*r = 2*π*0.19 m = 1.19 mThe time needed to complete a revolution is 2.4 s, so the translational speed can be written as follows:[tex]v =\frac{\Delta s}{\Delta t} = \frac{1.19m}{2.4s} = 0.5 m/s (1)[/tex]
The frequency in Hz is just the inverse of the time needed to complete a revolution (known as the period T), as follows:f = 1/T = 1/2.4s = 0.42 Hz (2)Finally, the angular speed is the rate of change of the angle rotated with respect to time, as follows:[tex]\omega = \frac{\Delta\theta}{\Delta t} = \frac{2*\pi}{2.4s} = 2.6 rad/sec (3)[/tex]
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, what is the magnitude of the force of kinetic friction (in N) acting on the crate
Answer:
The value is [tex]F_f = 46.935 \ N[/tex]
Explanation:
From the question we are told that
The magnitude of the horizontal force is [tex]F = 92.7 \ N[/tex]
The mass of the crate is [tex]m = 40.5 \ kg[/tex]
The acceleration of the crate is [tex]a = 1.13 \ m/s[/tex]
Generally the net force acting on the crate is mathematically represented as
[tex]F_{net} = F - F_f = ma[/tex]
Here [tex]F_f[/tex] is force of kinetic friction (in N) acting on the crate
So
[tex]92.7 - F_f = 40.5 * 1.13[/tex]
=> [tex]F_f = 46.935 \ N[/tex]
A teacher does a demonstration in front of a class. For the demonstration, a motion sensor is placed at the end of a long, horizontal track of negligible friction. A force sensor is used to pull a cart of known mass along the track, and the motion sensor tracks the motion of the cart. For several positions of the cart, a computer records the reading on the force sensor and the reading on the motion sensor for both the speed of the cart and the distance the cart moves away from the motion sensor. Based on only these data, which of the following scientific hypotheses could a student test?
a. The force exerted on the cart is directly proportional to the time the cart has moved.
b. The gravitational acceleration for an object on Earth is constant near Earth's surface.
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
d. As the force the sensor exerts on the cart increases, so does the acceleration of the cart.
e. Earth's gravitational force is a conservative force.
Answer:
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
Explanation:
In the context, a motion sensor detects the motion of the cart where a cart is used to move using a force sensor that is attached to the cart of certain mass. As the cart moves with a speed at a particular time, the computer records both the readings on the force sensor as well as the motion sensor. From these readings the is can be concluded that the work done on the cart is same as the change in the kinetic energy in the cart.
If the Earths atmosphere is air, then why doesn't the land fall? How is all the lava and water IN earth?
Answer:
Its stored in there
Explanation:
And air has little mass
You push a manual lawn mower across the lawn at constant speed. What is the value of the coefficient of friction between the mower and the grass
Answer:
0.27Explanation:
The question is incomplete. Here is the complete question:
You are pushing a 13.3 kg lawn mower across the lawn with a force of 200 N. What is the value of the coefficient of friction between the mower and the grass if the mower moves with a constant velocity? The force is applied downward at an angle of 65° with the horizontal.
According to Newton's second law of motion:
[tex]\sum F_x= ma_x\\F_{app} - F_f = ma_x\\[/tex]
[tex]F_f = \mu R\\[/tex]
[tex]F_{app} - \mu Rcos \theta = ma_x[/tex]
Fapp is the applied force = 200N
Ff is the frictional force
[tex]\mu[/tex] is the coefficient of friction between the mower and the grass
R is the reaction
m is the mass of the object
ax is the acceleration
Given
R = mg = 13.3*9.8
R = 130.34N
m = 13.3kg
ax = 0m/s² (constant velocity)
Fapp = 200N
[tex]\theta = 65^0[/tex]
Substitute the given parameters into the formula and get the coefficient of friction as shown;
Recall that: [tex]F_f = \mu R\\[/tex]
[tex]\mu = \frac{F_f}{R}\\\mu = \frac{F_{x}cos65}{F_y+W} \\\mu =\frac{ 200cos65}{200sin65+13.3(9.8)}\\\mu = \frac{84.52}{181.26+130.34}\\\mu = \frac{84.52}{311.6}\\\mu = 0.27[/tex]
Hence the coefficient of friction between the mower and the grass is 0.27
how does exercise help us be more enthusaistic in life.
A baby elephant trots in a straight line along a river. The horizontal position of the elephant in meters over
time is shown below.
Position
(meters)
15
12+
9+
6-
3+
+
4
+
16
8
12
20
24
-3+
-6+
Time
(seconds)
.92
Answer:
Displacement -6
Distance 24
Explanation:
Answer:
Explanation:
What is the displacement of the elephant between
0s
and
16s
displacement =9
distance traveled by the elephant between =9
Velocity ratio of single movable pulley is 2 when Velocity ratio of single fixed pulley is 1. why
Answer:
Velocity ratio of single pulley = 2
Explanation:
Given:
Tension T
Distance = T + T = 2T
Find:
Velocity ratio of single pulley
Computation:
Velocity ratio of single pulley = Total distance / Distance effort
Velocity ratio of single pulley = 2T / T
Velocity ratio of single pulley = 2
g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Answer:
339.3 N
Explanation:
First, we start by converting the units.
1 rev/s = 2π rad/s, so
0.6 rev/s = 2π * 0.6 rad/s
0.6 rev/s = 1.2π rad/s
0.6 rev/s = 3.77 rad/s
Now we apply the equation of motion,
W(f) = w(o) + αt
3.77 = 0 + α * 2
3.77 = 2α
α = 3.77/2
α = 1.885 rad/s²
Torque = I * α
Torque = F * r
This means that
I * α = F * r, where I = 1/2mr²
Substituting for I, we have
1/2mr²α = F * r, making F the subject of formula, we have
F = 1/2mrα, then we substitute for the values
F = 1/2 * 240 * 1.5 * 1.885
F = 678.6 / 2
F = 339.3 N
Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground, the more massive ball has ______ the kinetic energy of the less massive ball. (Neglect air friction.)
Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as
shown in Figure 1. Tarzan grabs a large bunch of bananas (15.0 kg) and grabs a vine in an attempt to swing up to his monkey who is 3.00 m above him as in Figure 2.
i. What is Tarzan’s momentum before he grabs the bananas? [2 points]
ii. What is Tarzan’s speed just after he grabs the bananas? [4 points]
iii. Can Tarzan swing high enough to reach his monkey? Justify your answer. [2 points]
We are given:
Mass of Tarzan before swinging = 80 kg
Mass of Tarzan when swinging = 80 + 15 = 95 kg
Velocity of Tarzan = 7 m/s
The height of the rock Tarzan's monkey is sitting on = 3 m
__________________________________________________________
Momentum of Tarzan before swinging:
We know that:
Momentum = Mass*Velocity
Momentum = 80 * 7
Momentum = 560 kg m/s
__________________________________________________________
Speed of Tarzan after grabbing the bananas:
The momentum of Tarzan will remain the same but his mass will increase
So, Since Momentum = New Mass* velocity
560 = 95 * v [where v is the velocity of Tarzan]
v = 5.9 m/s
__________________________________________________________
Finding the Initial and Final KE and PE:Here, KE = Kinetic Energy and PE = Potential Energy
Initial and Final KE:
We know that KE = 1/2*(mv²)
Initial KE:
Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]
Initial KE = 1/2*(95*5.9²)
Initial KE = 1653.5 Joules
Final KE:
Final KE = 1/2*(mv²)
[where v is the velocity at the maximum point of the swing]
Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s
Final KE = 1/2*(95*0²)
Final KE = 0 Joules
Initial and Final PE:
We know that:
PE = mgh
[where g is the acceleration due to gravity and h is the height]
Initial PE:
Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0
Initial PE = 95*10*0
Initial PE = 0 Joules
Final PE:
Let the maximum height of Tarzan be h m
Final PE = 95*10*h
Final PE = 950(h)
__________________________________________________________
Finding the maximum height Tarzan will reach:Here, KE = Kinetic Energy and PE = Potential Energy
From the law of conservation of momentum, we know that:
Initial KE + Initial PE = Final KE + Final PE
Replacing the variables:
1653.5 + 0 = 0 + 950h
1653.5 = 950h
h = 1653.5/950 [dividing both sides by 950]
h = 1.74 m
Therefore, the maximum height reached by Tarzan is 1.74 m
but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey