Answer:
I think its B. James would have a smaller mass on the Moon than he does on Earth.
sorry if i did it wrong
Can someone please write a summery about Yes day the movie from 2021 and it has to have something about Hygiene in it and this is for 100 points so can you please help me I will give you $100 and 100 points for a 2 page essay
Answer: This book is about a mom that is always saying no to her kids and is always pushing them to focus on what they are doing, but one day the mom or dad runs into a man that does what he calls "A Yes Day" with his kids and what this is, is for 24 hours you say yes to whatever you want to do and you aren't able to say no. So the mom tries yes day with her kids and they go through many crazy/extreme events. This has to deal with hygiene because the kids trash the house and get very dirty in a few parts of the movie. They always try to be as clean as possible, but the limits have been broken... This is why I think this movie has to deal with hygiene.
Answer:
I can't do 2 pages but I'll try
Explanation:
Yes day is a movie about parents and kids who usually fought, until someone suggested a yes day a day when parents can't say no. In the beginning of the movie, you'll first see that the parents used to say yes to everything, until they has kids they started saying no to everything. When someone suggested yes day they had to earn it, they did all the chores and they got a yes day. They made a list of 5 things they wanna do during yes day. i don't know what else to put for hygiene but yea.
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
what makes a funnel appear black
Answer:
Air
Explanation:
The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.
3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is
swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?
Answer:
the velocity of the big fish after the launch is 6 m/s.
Explanation:
Given;
mass of the big fish, m₁ = 5 kg
velocity of the big fish, u₁ = 8 m/s
mass of the small fish, m₂ = 1 kg
velocity of the small fish, u₂ = -4 m/s
Let the final velocity of the big fish after launch = v
Apply the principle of conservation linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
5 x 8 + 1 x (-4) = v(5 + 1)
40 - 4 = 6v
36 = 6v
v = 36/6
v = 6 m/s.
Therefore, the velocity of the big fish after the launch is 6 m/s.
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow
Answer:
132
Explanation:
if you round it is correct
Define threshing?
Brainliest For the right answer
Answer:
Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.
Answer:
Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.
The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.
In old times, grain had to be beaten out of the grass on a Threshing Floor.
Define the average acceleration of a particle
between two given instants.
Object A is negatively charged. Object A and Object B
attract. Object B and Object C repel. Object C and Object
D repel. What type of charge does Object B, Object C, and
Object D possess?
Answer:
Malrpr00qpq9owoowopwiaahaulaqkkkala9asoLHahababajjajalls
Explanation:
hhoootyiñlf7ogffyiklmhf
helllllpppppppppppp.
pleaseeeeeeee
Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:[tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]
Now, we need first to find the value of the angular acceleration, that we can get from the following expression:[tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]
Since the machine starts from rest, ω₀ = 0.We know the value of ωf₁ (the operating speed) in rev/min.Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:[tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]
Replacing by the givens in (2):[tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]
Solving for α:[tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]
Replacing (5) and Δt in (1), we get:[tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]
in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:[tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]
In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:[tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]
First of all, we need to find the value of the angular acceleration during the second period.We can use again (2) replacing by the givens:ωf =0 (the machine finally comes to an stop) ω₀ = ωf₁ = 57.5 rev/secΔt = 32 s[tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]
Solving for α in (9), we get:[tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]
Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:[tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]
In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:[tex]\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)[/tex]Which term does this explain?
This is a non-mathematical explanation of how nature works. It must be
supported by a large body of evidence.
Fact
Law
Theory
Hypothesis
What conversion takes place in a motor?
A. An electric current into a magnetic field
B. Mechanical energy into electric energy
C. Electric energy into mechanical energy
D. A lower voltage into a higher voltage
Thank you!! I will mark brainliest!!
A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it maintains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer
at the bottom of the slide.
Answer:
the velocity of the swimmer at the bottom of the slide is 6.26 m/s
Explanation:
The computation of the velocity of the swimmer at the bottom of the slide is given below:
v = √2gh
= √2 × 9.8 × 2
= 6.26 m/s
Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s
The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change with reference to time is referred to as the velocity
What is velocity?The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
The velocity is found as;
[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]
Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.
To learn more about the velocity refer to the link ;
https://brainly.com/question/862972
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.2 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.3 mm and that is perpendicular to the direction of the field.
poste en français s’il vous plaît
Pesticides sprayed on the plants in fields become When they settle and drip onto the soil .
A. Nutrients
B.Pollutants
C.Acid rain
D.Ammonia
Answer:nutrients
Explanation:
Answer:
pollutants is the right answer.
To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.
A. Find the acceleration a of the particle.
B. Evaluate the integral W = integarvf,vi mudu.
Answer:
a) the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Explanation:
Given the data in the question;
force on particle F = ma
displacement s = x[tex]_f[/tex] - x[tex]_i[/tex]
work done on the particle W = Fs = mas
we know that; change in energy = work done { work energy theorem }
[tex]\frac{1}{2}[/tex]m( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = mas
[tex]\frac{1}{2}[/tex]( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = as
( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = 2as
a = ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
Therefore, the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) Evaluate the integral W = [tex]\int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W = \int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W =m[\frac{v^{2} }{2} ]^{vf}_{vi}[/tex]
W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Therefore, the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
PLEASE I NEED HELP CLICK ON THIS IMAGE
I believe you are incorrect. A weathered mountain would appear more jagged.
I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.
If this is incorrect, please, don't refrain to tell me.
effect of high pitch on humans
Answer:
High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.
What are the two main processes carried out by the excretory system?
Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2
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7. Copper can be coated on the surface of iron but not on silver? why
Answer:
Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.
Explanation:
Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose
. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder
Answer:
As we can see, a string is attached with block A, and three string is folded with ply which is attached with B
x
B
=3x
A
Now differentiate with respect to x
V
B
=3V
A
Given,
V
A
=0.6m/s(totheright)
So,
V
B
=0.6×3
=1.8m/s(downward)
Explanation:
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In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find
an electron's acceleration in this field.
Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
[tex]E = \frac{F}{q}\\\\F = Eq\\[/tex]
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\[/tex]
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
[tex]F = ma\\a = \frac{F}{m}\\[/tex]
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
[tex]a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\[/tex]
a = 1.055 x 10¹⁷ m/s²
How do you find the period of a sound wave?
Answer:
Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.
Which disciplines were developed based on the Greek questioning of the elements?
A. art and music
B. chemistry and biology
C. tragedy and comedy
If a true bearing of a ship at sea is 227°, what is its direction angle?
A. 43°
B.313°
C. 223°