Answer:
cm/s
6
128.2
96.0
7
145.8
Table B
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?
Answer:
[tex]v=6.65m/sec[/tex]
Explanation:
From the Question we are told that:
Mass [tex]m=97.6[/tex]
Coefficient of kinetic friction [tex]\mu k=0.555[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\mu mg[/tex]
[tex]F=0.555*97.6*9.8[/tex]
[tex]F=531.388N[/tex]
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
[tex]a_f=-\mu g[/tex]
[tex]a_f=-0.555 *9.81[/tex]
[tex]a_f=-54455m/sec^2[/tex]
Therefore
[tex]v=u-at[/tex]
[tex]v=0+5.45*1.22[/tex]
[tex]v=6.65m/sec[/tex]
You are driving home from school steadily at 97 km/h for 190 km . It then begins to rain and you slow to 60 km/h instantly. You arrive home after driving 4.0 hours.
how far is your hometown from school?
Please delete my answer. I made a mistake
The motion that is not repeated in regular interval of time is called....?
please help me!
Answer:
Motion which repeats itself after regular intervals of time is known as periodic motion.Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3m3 of water flows through the dam each second. The water is released 220 mm below the top of the reservoir. If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?
Answer:
The output electric power is 1338876 W.
Explanation:
Volume, V = 690 cubic meter
height, h = 220 mm = 0.22 m
efficiency = 90 %
time , t = 1 s
Let the mass is m.
m = volume x density
m = 690 x 1000 = 690000 kg
The input power is
P = m g h = 690000 x 9.8 x 0.22 = 1497640 W
The electric power out put is
[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?
Explanation:
the answer is in the above image
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
If the temperature stays constant, which change would decrease the amount
of thermal energy in an object?
A. Decreasing its density
B. Increasing its velocity
c. Decreasing its mass
D. Increasing its mass
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
3) show thal escape veloerty ve where symbols have their usual meaning?
Answer:
[tex]v = \sqrt{\frac{2GM}{r} }[/tex]
The formula for escape velocity where:
G - Gravitational constant (9.81 etc.)
M - the mass of the object the escape should be made from
r - distance to the centre of that mass
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
pls mark me brainlest
Can someone help me
Btw the last one say current
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down? Assume the speed of sound is 343 m/s.
Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
Two stars that are in the same constellation:
Answer:
A binary star is a star system consisting of two stars orbiting around their common barycenter.
❣️(◍Jess bregoli◍)❣️#keep learning
To accurately describe the wind, the measurement should include
A) a direction, but not a speed
B)a speed, but not a direction
C) both a speed and a direction
D) neither a speed nor a direction
Answer:
C. both a speed and a direction
Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
The object has a mass of 100kg. The Tension is 200N[U]. What is the acceleration of this elevator? *
A) 8m/s/s
B) 8m/s/s[D]
C) 9.8m/s/s[D]
D) 0.5m/s/s[D]
Answer:
So the answer is B. A is wrong because negative answer = deceleration
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system
Complete Question
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.
(a) What is the moment of inertia of the system
(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm
Answer:
a) [tex]M.I=0.32kg.m^2[/tex]
b) [tex]K.E=621.8J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=2.0kg[/tex]
Disk Radius [tex]r_d=50cm=0.5m[/tex]
Mass of annular cylinder [tex]M_c=1.0kg[/tex]
Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]
Outer Radius of cylinder [tex]R_o=0.3m[/tex]
Angular Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]
[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]
[tex]M.I=0.32kg.m^2[/tex]
Generally the equation for Rotational Kinetic Energy is mathematically given by
[tex]K.E=0.5M.I \omega^2[/tex]
[tex]K.E=0.5 *0.32*62.83[/tex]
[tex]K.E=621.8J[/tex]
The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.
Answer:
Option B. 3.25×10¯¹⁰ mm.
Explanation:
Measurement (cm) = 0.00325×10⁻⁸ cm
Measurement (mm) =?
The measurement in mm can be obtained as follow:
1 cm = 10 mm
Therefore,
0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm
0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm
Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.
The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm
The number given is in standard form and can be written as 3.25*10^-11 cm.
To convert this from centimeter to millimeter, we have to multiply this value by 10.
Conversion Units1 cm - 10mm100cm = 1m1000m = 1kmSo, let's 3.25*10^-11 by 10 and get our value in mm
[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]
From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]
Learn more about conversion of units here;
https://brainly.com/question/8426032
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify
Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
Also read it;
https://brainly.com/question/11049671
#SPJ2
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond. One skater throws a 0.4-kg ball at 6 m/s to his friend, who catches it and throws it back at 6.0 m/s. When the first skater has caught the returned ball, what is the velocity of each of the two skaters
Answer:
v = 0.064 m/s
Explanation:
Given that,
The mass of two skaters = 75 kg
The mass of a ball = 0.4 kg
The speed of the ball = 6 m/s
The speed of skater = 6 m/s
We need to find the velocity of each of the two skaters.
Under the values given the moment with respect to the ball and which is subsequently transmitted to people it would be given by:
[tex]P=0.4(6)+0.4(6)\\\\=4.8\ kg-m/s[/tex]
We know that,
P = mv
Where
v is the velocity of each skater.
[tex]v=\dfrac{p}{m}\\\\v=\dfrac{4.8}{75}\\\\=0.064\ m/s[/tex]
So, the velocity of each of the skaters is 0.064 m/s.
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
