Answer:
0.75km/min
Explanation:
I haven't learnt physics at all but I'll try
57336 - 57321 = 15km (Difference)
8.50 am - 8.30 am = 20 minutes (Difference)
If the car odometer adds up 15km every 20 minutes...
20 mins = 15km
1 min = 15 ÷ 20 = 0.75km
How much does a crate weigh (force) if it is 0.25 meters wide and 1.5 meters long and it
exerts 800 pascals of pressure on the ground when it lands?
Force of a crate : 300 N
Further explanationGiven
0.25 meters wide and 1.5 meters long
Pressure = 800 Pa
Required
Force
Solution
P = F/A
A = 0.25 x 1.5
A = 0.375 m²
P = 800 Pa = 800 N/m²
Input the value :
F = P x A
F = 800 N/m² x 0.375 m²
F = 300 N
How does Boyle's laws describe the relationship between gas pressure and volume?
Answer:
For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. Or Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship. If volume increases, then pressure decreases and vice versa, when the temperature is held constant.
What part of the eye protects your eye from scratches?
retina
pupil
cornea
iris
Answer:
Cornea
Explanation:
The outer layer of the cornea is covered by epithelial cells that protect the surface. A corneal abrasion may result from dragging a piece of debris across the surface and scraping off the epithelial cells.
a bear scratches his claws on a big pine tree. Does the bear do work on the tree? why or why not?
(btw it has to do with physics) ITS NOT A GENERAL QUESTION.
Answer:
See the explanation below.
Explanation:
Work in physics is defined as the product of force by the distance that the body travels in the direction of the force.
It can be represented by means of the following equation.
[tex]W=F*d[/tex]
where:
W = work [J]
F = force [N]
d = distance [m]
In the given example the work is zero since the tree does not move, therefore the bear exerts a force on the tree. But there is no talk of movement of the tree, therefore the work is zero.
Read through the and calculate the predicted change in kinetic energy of the oblect compared to 50 kg ball traveling at 10 m/s . 50 kg ball traveling at 20 n / s would havekinetic energy 50 kg traveling at 5 m/s would have energy 50 kg person falling at 10 m/s would havekinetic energy
Answer:
A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.
A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.
A 50 kg person falling at 10 m/s would have the same kinetic energy.
Explanation:
hope this helps:)
what is electric switch
Answer:
A device that is used to break an electric circuit is called electric switch
A boat travels 90 m in 10 seconds, what is its speed?
90 m/s
9 mi/hr
9 m/s
900 m/s
Answer:
it is 9m/s , hope it helps
A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The final velicoty of the bowling pin is _____m/s.
Answer: hope this helps
Explanation:
If the final velocity of the ball is 2.5 m/s, the final velocity of the bowling pin is 7.8125 m/s.
What is principle of conservation of momentum?According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.
According to the principle of momentum conservation, total momentum of an system remains conserved.
Total initial momentum = total final momentum
5 kg × 5 m/s + 1.6 kg × 0 m/s = 5 kg × 2.5 m/s + 1.6 kg × v
1.6 kg × v = 5 kg × 5 m/s - 5 kg × 2.5 m/s
v = (5 kg /1.6 kg) (5 m/s - 2.5 m/s)
= 7.8125 m/s
Hence, If the final velocity of the ball is 2.5 m/s, the final velocity of the bowling pin is 7.8125 m/s.
Learn more about momentum here:
brainly.com/question/24030570
#SPJ5
A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working. The student goes to the kitchen and prepares 10 of 100°C water to mix with 10 L of 10°C water.
Answer:
(10 L)(100° C) + (90 L)(10°) = (100 L)T
1,000 + 900 = 100T
1,900 = 100T
100T = 1,900
T = 1,900/100
Explanation:
Which examples best demonstrate likely tasks for Legal Services workers? Check all that apply.
Cisco inspects an office building to make sure it meets fire safety regulations.
Gina gathers information about a court case.
Saul oversees inmates at a correctional facility.
Hana interviews and advises a person who has been accused of a crime.
Pamela pursues and arrests a person suspected of a crime.
Dewayne creates the paperwork for a business contract.
Answer:
b. d. f.
Explanation:
Got it right on Edge
Answer:
2. Gina gathers information about a court case.
4. Hana interviews and advises a person who has been accused of a crime.
6. Dewayne creates the paperwork for a business contract.
PLEASE HELP ASAP!!!!
Sort the statements based on whether they are correct or incorrect.
All waves have the same
wavelength.
Radio telescopes give us
information about the
birthplace of a star.
If the wavelength of a wave
increases, its frequency
also increases.
Visible light is a kind of
electromagnetic wave.
All objects emit
electromagnetic waves.
Answer: the first is incorrect the second is correct the third is correct
Explanation: the fourth is correct the fifth is incorrect
98. In Fig. 24-71, a metal sphere
with charge q = 5.00 uC and radius
r = 3.00 cm is concentric with a
larger metal sphere with charge Q =
15.0 uC and radius R = 6.00 cm. (a)
What is the potential difference be-
tween the spheres? If we connect the
spheres with a wire, what then is the
charge on (b) the smaller sphere and
(c) the larger sphere?
Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6.[tex]\overline 6[/tex] μC
(c) The charge on the smaller sphere, Q₁ = 13.[tex]\overline 3[/tex] μC
Explanation:
(a) The given parameters are;
The charge on the inner sphere, q = 5.00 μC
The radius of the inner sphere, r = 3.00 cm = 0.03 m
The charge on the larger sphere, Q = 15.0 μμC
The radius of the larger sphere, R = 6.00 cm = 0.06 m
The potential difference between two concentric spheres is given according to the following equation;
[tex]V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)[/tex]
Where;
R = The radius of the larger sphere = 0.06 m
r = The radius of the inner sphere = 0.03 m
q = The charge of the inner sphere = 5.00 × 10⁻⁶ C
Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C
k = 9 × 10⁹ N·m²/C²
Therefore, by plugging in the value of the variables, we have;
[tex]V_r - V_R = 9 \times 10^9 \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000[/tex]
The potential difference between the spheres, [tex]V_r - V_R[/tex] = 750,000 N·m/C = 750 kVA
(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;
[tex]Q_f[/tex] = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
[tex]Q_f[/tex] = 20 × 10⁻⁶ C
From which we have;
Q₁/Q₂ = R/r
Where;
Q₁ = The new charge on the on the larger sphere
Q₂ = The new charge on the on the smaller sphere
[tex]Q_f[/tex] = 20 × 10⁻⁶ C = Q₁ + Q₂
∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂
∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2
20 μC - Q₂ = 2·Q₂
20 μC = 3·Q₂
Q₂ = 20 μC/3
The charge on the smaller sphere, Q₂ = 20 μC/3 = 6.[tex]\overline 6[/tex] μC
(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3
The charge on the smaller sphere, Q₁ = 40 μC/3 = 13.[tex]\overline 3[/tex] μC.
The potential difference is 7.5*10^5V and the charge on the smaller sphere is 6.67uC while the charge on the larger sphere is 13.34uC
Data;
q = 5.0 uCr = 3.0cm= 0.03mQ = 15.0uCR = 6.0cm = 0.06mPotential Difference between the Spheresfor the inner sphere;
[tex]v_i = \frac{kq}{r} = \frac{(9*10^9)(5*10^-^6)}{0.03} \\v_i = 1.5*10^6V[/tex]
for the outer sphere;
[tex]v_o = \frac{KQ}{R} = \frac{(9*10^9)(15*10^-^6)}{0.06} = 2.25 * 10^6V[/tex]
The difference in potential is
[tex]\delta V = v_o - v_i = 2.25*10^6 - 1.5*10^6 = 7.5*10^5V[/tex]
Charge on the Smaller Plate[tex]q + Q = 5 + 15 = 20 \mu C = 20*10^-^6C ..eq(i)[/tex]
The sharing of charge continues till they attain a point of equal potential
[tex]v_i = v_o \\\frac{kq}{r} = \frac{kQ}{R} \\\frac{q}{0.03} = \frac{Q}{0.06} \\Q= 2q ...eq(ii)\\[/tex]
let's solve for equation (i) and equation (ii)
[tex]q+2q = 20\mu C\\q = 6.67 \mu C[/tex]
The charge on the smaller sphere is 6.67uC
The charge on the larger sphereThe charge on the larger sphere is
[tex]Q = 2q\\q = 6.67 \mu C\\Q = 2 * 6.67 = 13.34 \mu C[/tex]
The charge on the larger sphere is 13.34uC
Learn more on potential difference on a charged particle here;
https://brainly.com/question/9383604
which of the following cannot be used for measurement of time
a A leaking tap
b simple pendulum
c shadow of object during the day
d blinking of eyes
Answer:
D. blinking ofeyea
Explanation:
not really sure
Which one of the following is NOT a phenotype?
O Red Hair
O Brown Eyes
O Red/Green Blood
O Five foot tall girl
O These are all phenotypes
What force will accelerate a 20 kg object at 4 m/s 2? *
Answer:
80 N
Explanation:
mass = 20 kg
acceleration = 4 m/s^2
Force = mass * acceleration
= 20 *4
= 80 N
A man can row a boat at 4kmhr in still water. He rows the boat 2km upstream and 2 km back to his starting place in 2 hours. How fast is the stream flowing
Answer:
2.83km/hr
Explanation:
Let stream flowing at the rat x km/hr
In still water
Speed of boat=4km/hr
Upstream speed=4-x
Downstream speed=4+x
Total time=2 hr
Time=Distance/speed
According to question
[tex]\frac{2}{4+x}+\frac{2}{4-x}=2[/tex]
[tex]\frac{8-2x+8+2x}{(4+x)(4-x)}=2[/tex]
[tex]\frac{16}{4^2-x^2}=2[/tex]
Using the formula
[tex](a+b)(a-b)=a^2-b^2[/tex]
[tex]4^2-x^2=\frac{16}{2}[/tex]
[tex]16-x^2=8[/tex]
[tex]16-8=x^2[/tex]
[tex]8=x^2[/tex]
[tex]x=\sqrt{8}=2.83km/hr[/tex]
Hence, the stream is flowing at 2.83km/hr
Need help ASAP
Will mark BRAINLIST and thanks only for correct answers
1.What is the correct way a cell phone call goes out?
-The phone call goes out, transmits through microwaves, then moves to the central hub. Then the signals go to the nearest tower to the receiver of the call.
-The phone call goes out, transmits through radio waves, then moves to the central hub. Then the signals go to the nearest tower to the receiver of the call.
-The phone call goes out, transmits through microwaves, then moves to the tower. Then the signals go to the nearest hub to the receiver of the call.
2.Which property of AM frequencies allow them to travel over long distances?
-high amplitude
-high energy
-constant velocity
-long wavelength
3.what types of waves do cell phones use?
-radio waves
-gamma rays
-microwaves
Answer:
1. The correct way a cell phone call goes out is;
The phone call goes out, transmits through radio waves, then moves to the central hub. Then the signal go to the nearest tower to the receiver of the call
2. Long wavelength
3. Microwave
Explanation:
1. When making a call through a cell phone, the speech made is digitized such that it is converted to electrical signals which are then converted into number strings that are transmitted out of the antenna of the phone as radio waves which are then picked up by the closest cellphone tower mast which then transmit the signal to the applicable central hub from where the signal is transmitted to the tower closest to the intended call receiver
Therefore, the correct option is that the phone call goes out, transmits through radio waves, then moves to the central hub. Then the signal go to the nearest tower to the receiver of the call
2. The property of AM frequencies that allow them to travel over long distances is long wavelength
The frequency of AM radio waves is between 540 kHz and 1600 kHz, which can be propagated as ground wave or by the ionosphere at night
3. Cell phones makes use of the subcategory of microwave which is the super high frequency (SHF) band which with a range of 3 GHz to 30 GHz
Need help ASAP
Thanks + BRAINLIST only for correct answer
1.Cell phones convert sound waves into______waves.
-light waves
-sound waves
-radio waves
2.Electromagnetic waves used in cell phone communications are called?
-microwave
-radio wave
-infrared
-all of the above
3.To send out a radio signal far and wide...it is called?
-broadcast
-casting
-antenna
Answer:
.Cell phones convert sound waves into______waves.
-sound waves
.Electromagnetic waves used in cell phone communications are called?
-radio wave
.To send out a radio signal far and wide...it is called?
antenna
Explanation:
i might be wrong in ans c ok
how find \empirical formula
Pleaseeeeeeeeeeeeeeeeeee
i’m stuck in B and D. which one is it?
Answer:
Option b is correct - Rocky, nearly continuous, found on earth surface.
The reason is it is the rocky surface area composed of crust and upper mantle part. Basically, it is the outer shell of earth.
how do you fix a broken foot
my sister is in literal tears and i dont have a phone and
our parents are out of the house rn
Please Help Fast Select the correct answer. Which equation correctly relates kinetic energy, mass, and velocity? A. B. C. D.
Answer:
KE = 1/2 m v^2
Explanation:
Determine the momentum of a system that consists of two objects. One object, m1, has a mass of 6 kg and a velocity of 13 m/s towards the east and a second object, m2, has a mass of 14 kg and a velocity of 7 m/s in that same direction.
Momentum = (mass) x (velocity)
Momentum m₁ = (6 kg) x (13 m/s east) = 78 kg-m/s east
Momentum m₂ = (14 kg) x (7 m/s east) = 98 kg-m/s east
Since they're both in exactly the same direction you can simply addum up to get their total momentum. It's 176 kg-m/s east .
The momentum of a system that consists of two objects should be 176 kg m/s on east hand side.
Calculation of the momentum:Since we know that
Momentum = (mass) x (velocity)
So,
Momentum m₁ = (6 kg) x (13 m/s east) = 78 kg-m/s east
Momentum m₂ = (14 kg) x (7 m/s east) = 98 kg-m/s east
Since they are on same direction so the total momentum should be
= 78 + 98
= 176 kg-m/s east
hence, The momentum of a system that consists of two objects should be 176 kg m/s on east hand side.
Learn more about momentum here: https://brainly.com/question/24815698
what are the laws of motion
James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the
air from. Before he removes the air, he can hear the alarm ringing. After he removes the alr, he cannot hear the
alarm ringing. What condusion can you draw from this experiment?
A. Sound waves can travel in a vacuum
B. Sound waves are transverse waves and are loud
C. Sound waves cannot travel through a medium
D. Sound waves require a medium to travel through
Answer: c
Explanation:
Sound waves cannot travel through a medium
Part C
How many combined bits of storage are there on the two tablets?
Answer:
IVE BEEN TRYNA FIND THE ANSWER FOR THE PAST HOUR T-T
Explanation:
Answer:
3.2 x10^11 bits of storage
Explanation:
A stack of 10 quarters weighs 0.1255 lb. what is the mass in grams of one quarter?
Answer:
one quarter= 5.69 g
Explanation:
if 10 quarters = 0.1255 lbs
then 1 quarter = 0.01255 lbs (0.1255 divided by 10)
0.01255 lbs = 5.69 g
what is gas pressure
A body at rest is given an instal
uniform acceleration of 8.0m/s² for 30
seconds after which the acceleration is
reduced at 50m/s² for the next 20
second the body maintains the speed
attain for 60 seconds after which it is
brought to rest in 20 seconds
v=u+at
v=0+8.0×30
v=240