Answer:
D) Medical School
Explanation:
Edge 2021
A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.
a. How long does it take to reach the ground?
b. How far does the rock land from the base of the cliff?
Answer:
a. t = 2.02 s
b. d = 20.2 m
Explanation:
Horizontal Motion
If an object is thrown horizontally from a height h with a speed v, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The time the object takes to hit the ground can be calculated as follows:
[tex]\displaystyle t=\sqrt{\frac{2h}{g}}[/tex]
The time does not depend on the initial speed.
The range or maximum horizontal distance traveled by the object can be calculated by the equation:
[tex]\displaystyle d=v.t[/tex]
The man standing on the edge of the h=20 m cliff throws a rock with an initial horizontal speed of v=10 m/s.
a.
The time taken by the rock to reach the ground is:
[tex]\displaystyle t=\sqrt{\frac{2*20}{9.8}}[/tex]
[tex]\displaystyle t=\sqrt{4.0816}[/tex]
t = 2.02 s
b.
The range is:
[tex]\displaystyle d=10\cdot 2.02[/tex]
d = 20.2 m
Solve ASAP please!!!
Which of these best describes science?
A-Science is the study of fundamental problems such as existence, values, and reason.
B-Science is the collection of all living things on Earth.
C-Science is a body of knowledge and a process of discovery that allows people to understand the natural
world.
D-Science is the practice of diagnosing, treating, and preventing disease.
What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s
Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...
Hope this helps....... Stay safe and have a Merry Christmas!!!!!!!!!! :D
Write true if the statement is correct and false if the statement
1. Magnetic poles always appear in pair.
2. Iron and nickel are examples of magnetic materials.
3. The two ends of a magnet are called magnetic poles.
4. Two similar magnetic poles attract each other
5. If one end of an iron bar attracts one pole of a com
Grade 7 Physics
then the iron bar must be a magnet.
Match the terms in column A with the phrases in colu
B
Answer:
true false
Explanation:
Answer:
1.true
2.true
3.true 4.false 5.true
Place the molecules in the list in order with the smallest molecules at the top and largest molecules at the bottom: starch molecule, water molecule, glucose molecule, carbon dioxide molecule, protein molecule, amino acid molecule, oxygen molecule
Explanation:
list in order with the smallest molecules at the top and largest molecules at the bottom:
1.Oxygen
2.Water
3.Carbon dioxide
4.amino acid
5.Glucose
6. Starch
7. Protein
The molecules can be arranged as follows with the smallest at the top is oxygen, water, carbon dioxide, amino acid, glucose, starch, protein.
The given molecules;
starch, water, glucose, carbon dioxide, protein, amino acid, oxygen
The molecular weight of the given compounds is listed as follows;
water -------- 18 g/mol
glucose ------- 180.2 g/mol
starch (2 or more glucose) ------ 360 g/mol
carbon dioxide ------ 44 g/mol
protein -------- 20,000 g/mol
amino acid (glycine) ----- 75 g/mol
oxygen -------- 16 g/mol
Thus, the molecules can be arranged as follows with the smallest at the top;
oxygenwatercarbon dioxideamino acidglucosestarchproteinLearn more here:https://brainly.com/question/19537339
please answer it fast and correctly
Answer:
(1) chess is played by him
(2) The great black engine was seen by her
(3) Arrangements for functions had been made by him
(4) A meeting is being attended by Raman
= when the last of the guests left, I went back into the hall to look for my mobile phone that I had kept on the table, I found that my mobile phone was not there. I felt freaked and I asked to every member of my family about my phone, but nobody had idea about that. then there was a phone call made by my uncle to my father, that his son had taken the mobile by mistake. After sometime, My uncle came to return my mobile. Then I took a sigh of relief .
please mark me as a brainelist
what is the correct answer?
Answer:
3.2
Explanation:
I remember taking this quiz and I think that one is correct
Ramas weight is 40 kg. She is carrying a load of 20 kg up to a height of 20 m. What work does she do? What work does she do?
Answer:
The work done by Ramas is 3920 N.
Explanation:
Given;
mass of the load lifted by Ramas, m = 20 kg
height through which the load is lifted, h = 20 m
The work done by Ramas is equal to gravitational potential due to the height in which the load is lifted.
W = PE = mgh
where;
g is acceleration due to gravity = 9.8 m/s²
W = 20 x 9.8 x 20
W = 3920 N.
Therefore, the work done by Ramas is 3920 N.
A 0.05 kg golf ball at rest was hit with a force of 200N. The golf ball left the club at an angle of 35° to the horizontal at 49m/s. What was the final velocity as it hits the ground
Note that when the ball is in the air, the only force acting on it is the downward pull of gravity, so the force of the swing is a red herring.
The ball hits the ground at the same height from which it was hit, so its final velocity will have the same magnitude of 49 m/s, but pointing at an angle of 35° below the horizontal.
Uncle Harry weighs 750 N. What is his mass in kg?
Answer:
W = MG
750 = M * 10
M = 750/10
M = 75 kg
Answer:
76.479 Kilograms Force (kgf)
Explanation:
1. What is the speed of a runner who runs 400 meters in 40
seconds?
Answer:
10 m/s
Explanation:
Speed = Distance / Time
Speed = 400 / 40
Speed = 10
Unit = meters per second = m/s
A 6 kg block is released from rest at the top of an incline, as shown above, and slides to the bottom. The incline is 1.97 m long, is inclined at an angle of 29.4°, and the coefficient of kinetic friction between the block and the incline is 0.21.
a. Calculate the normal force the incline puts on the block.
The net force on the block acting perpendicular to the incline is
∑ F = n - w cos(29.4°) = 0
where n is the magnitude of the normal force and w = m g is the weight of the block.
The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.
Solve for n :
n = m g cos(29.4°)
n = (6 kg) (9.80 m/s²) cos(29.4°)
n ≈ 51.2 N
Which of the following are single-displacement reactions?
A Mg (s) + 2H,0 (g) → Mg(OH)2 (aq) + Hz (g)
B. 2C,H6 (g) + 70, (g) → 6H,0 (9) + 4CO, (g)
c. AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaN3 (aq)
D. Clz (9) + KBr (aq) → KCI (aq) + Br2 ()
Multiple answer are allowed!!!
Answer:
A & D
Explanation:
A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.
From the options, only options A & D fits this definition of single-displacement reactions.
For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.
For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.
The other options are not correct because they don't involve only and element and a compound on each side of the reaction.
A string of length 0.67 m and mass 2.0 g is held at both ends under tension. The string
produces a 700 Hz tone when it vibrates in the fourth harmonic. During this event, what is the
wavelength of the standing wave in the string?
.34
.55
.61
.45
.49
Answer:
The value is [tex]\lambda = 0.335 \ m[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 0.67 \ m[/tex]
The mass is [tex]m = 2.0 \ g = 0.002 \ kg[/tex]
The frequency produced at fourth harmonic vibration(i.e n =4 ) is [tex]f = 700 \ Hz[/tex]
Generally the wavelength of the string is mathematically represented as
[tex]\lambda = \frac{ 2L }{n }[/tex]
=> [tex]\lambda = \frac{ 2 * 0.67 }{ 4 }[/tex]
=> [tex]\lambda = 0.335 \ m[/tex]
difference between lower fixed point and upper fixed point
Answer:
The lower fixed point, or ice point, is the temperature of pure melting ice at normal atmospheric pressure. The upper fixed point, or steam point, is the temperature of pure boiling water at normal atmospheric pressure.
Explanation:
Un cuerpo se lanza hacia arriba con 50 m/s . ¿A qué altura se encuentra cuando su rapidez es 20 m/s por segunda vez? (g = 10 m/s ^ 2) .
Answer:
105 m
Explanation:
Given that the initial velocity of the body, u= 50 m/s in upward direction.
Acceleration due to gravity, g= 10 m / [tex]s ^ 2[/tex] in the downward direction.
Let h be the height of the body when the velocity of the body is 20 m/s.
As the gravitational force acting in the downward direction, so taking it negative as
a= - 10 m/[tex]s ^ 2[/tex] in the upward direction.
By using the equation of motion [tex]v^2=u^2+2as[/tex] ...(i)
Here, u= 50 m/s
v= 20 m/s
[tex]a=- 10 m/s^2[/tex]
s=h
Putting all the values in the equation (i), we have
[tex]20^2=50^2+2(-10)h \\\\400=2500-20h \\\\h=\frac{2500-400}{20} \\\\h=105 m[/tex]
Hence, the height of the body is 105 m.
An object has a momentum of 56.59 kg*m/s and a velocity of 5.3 m/s. What 1 point
is the mass of the object? *
Answer:
10.68 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{p}{v} \\ [/tex]
p is the momentum
v is the velocity
From the question we have
[tex]m = \frac{56.59}{5.3} \\ = 10.677358...[/tex]
We have the final answer as
10.68 kgHope this helps you
A cello string is 0.695 m long, and transmits waves at 204 m/s. What frequency does it produce? (Unit = Hz)
Answer:
ν=293.53 Hz
Explanation:
ν= speed/λ
speed = 204 m/s
λ=0.695 m
v = 204 m/s * 1/0.695m
meters cancels out
1/s = Hz
v = 204/0.695 Hz
ν=293.53 Hz
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apparently it is wrong, but my work and online calculator says otherwise... possibly round your answer? I'm not sure then...
Answer:146.75
Explanation:
What is an astronomical unit or AU
Answer:
D
Explanation:
Answer:
Astronomical Unit = 4.8481×10−6 pc; 1.5813×1.
Or even AU = Equal to
Explanation:
A communication satellite is in a circular path orbit around Earth. If the speed of the satellite is constant, the net force acting on the satellite ____.
Answer:
is changing in direction, but constant in magnitude
Explanation:
This question is a bit tricky since the velocity of the satellite is changing, but the speed is constant.
Speed is simply a measure of how fast you are going. It doesn't matter where you're going, just how quickly.
Velocity, on the other hand, does care about which direction you're going. For example, it could be then when you travel right, your velocity is positive, and when you travel left, your velocity is negative. This is the similar for a 2D shape like a circular orbit
Since we know velocity is changing, there must be acceleration which changes that velocity (since acceleration is the change in velocity: going from 0 to 60 mph, for example)
Thus, with a non-zero net acceleration, we know that there must be a force that is changing in direction, but constant in magnitude (since the orbit is a circle, and always attracted to the center of the Earth at equal distance).
When a 54.9-g tennis ball is served, it accelerates from rest to a speed of 45.1 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 35.1 cm. What is the magnitude of the net force acting on the ball
Answer:
F = 159.07 N
Explanation:
Given that,
Mass of a tennis ball, m = 54.9 g = 0.0549 kg
It accelerates from rest to a speed of 45.1 m/s.
The impact with the racket gives the ball a constant acceleration over a distance of 35.1 cm = 0.351 m
We need to find the magnitude of net force acting on the ball.
Let a be the acceleration of the ball. Using third equation of motion to find it.
[tex]v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(45.1)^2-(0)^2}{2\times 0.351}\\\\=2897.45\ m/s^2[/tex]
Net force,
F = ma
F = 0.0549 kg × 2897.45 m/s²
= 159.07 N
Hence, the net force acting on the ball is 159.07 N.
Calculate the gravitational force between a 4.0 x 10^8 kg mass and a 2.8 x 10^5 kg mass separated by a distance of 1.0 x 10^4 m.
Answer:
[tex]F_{g}=7.4704\times10^{-5}[/tex]
Explanation:
Givens: d = 1.0 × [tex]10^{4}[/tex] [tex]m_{1} = 4.0 x 10^{8}[/tex] [tex]m_{2} = 2.8[/tex]×[tex]10^{5}[/tex]
Required: [tex]F_{g}=?[/tex]
Formula: [tex]F_{g}= \frac{(6.67x10^{-11})(m_{1})(m_{2}) }{d^{2}}[/tex] (6.67x10^-11 is the gravitational constant)
[tex]F_{g}=\frac{\left(6.67\cdot10^{-11}\right)\left(4.0\cdot10^{8}\right)\left(2.8\cdot10^{5}\right)}{\left(1.0\cdot10^{4}\right)^{2}}[/tex]
[tex]F_{g}=7.4704\times10^{-5}[/tex]
How much work is it to lift a 2.0 kg sack of potatoes vertically 6.5 m?
Answer:
Therefore, the work done to lift a 20 kg sack of potatoes of potatoes vertically 6.5 m is 1, 274 J.
If an object has a weight of 4050 N near the surface of the Earth, what is its mass?
Answer:
417.5 kg
Explanation:
Since the gravitational for of attraction F, equals the weight of the object W when its close to the earth, F = GMm/R² = W = mg where m = mass of object and g = acceleration due to gravity close to the earth = 9.8 m/s².
g = GM/R² where g = acceleration due to gravity close to the earth, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.927 × 10²⁴ kg and R = radius of earth = 6.4 × 10⁶ m
So, g = GM/R²
= 6.67 × 10⁻¹¹ Nm²/kg² × 5.927 × 10²⁴ kg/(6.4 × 10⁶ m)²
= 39.53 × 10¹³ Nm²/kg ÷ 40.96 × 10¹² m²
= 9.65 N/kg
≅ 9.7 N/kg
= 9.7 m/s²
Since W = 4050 N and W = mg
m = W/g = 4050 N/9.7 m/s²
= 417.53 kg
≅ 417.5 kg
So, the mass of the object is 417.5 kg
I WILL MARK BRAINLIEST!
A 2.0 kg bullet is fired into a sandbag sitting on a wagon. The combined mass of the wagon and the sandbag is 4.5 kg. a) If the wagon/sandbag/bullet moved together with a speed of 0.22 m/s immediately after impact with the bullet, with what speed did the bullet enter the sandbag? b) Is this an elastic or inelastic collision? Explain.
Answer:
To answer your question use the code ICE on here to get your answer works every time for me hope this helps
In an A.C circuit current leads voltage by phase π/2 then circuit is
Answer:
capacitive
Explanation:
In a capacitive circuit, current is proportional to the derivative of the voltage. For a sinusoidal excitation, this means current is at the highest level when voltage is increasing through zero. That is, current leads the voltage.
What is the relationship between the speed of impact and the dropping height?
A ship is travelling due east at 30 km/hr and a boy runs across the deck
in a south west direction at 10 km/hr. Find the velocity of the boy
relative to sea.
Answer:
Vr = 20 [km/h]
Explanation:
In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.
Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).
[tex]V_{r}=30-10\\V_{r}=20 [km/h][/tex]
the kinetic energy of an object of mass in moving with a velocity of 5 MS -1 is 25j what will be its kinetic energy when its velocity is doubled what will be its kinetic energy when its velocity is increased three times. please answer fast I will mark brainly
Answer:
When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J
Explanation:
Kinetic Energy
Is the type of energy an object has due to its speed. It's proportional to the square of the speed.
The equation for the kinetic energy is:
[tex]\displaystyle K=\frac{1}{2}mv^2[/tex]
Where:
m = mass of the object
v = speed at which the object moves
The kinetic energy is expressed in Joules (J)
The object has a kinetic energy of K=25 J when moving at v=5 m/s, thus the mass can be calculated by solving for m:
[tex]\displaystyle m=\frac{2K}{v^2}[/tex]
[tex]\displaystyle m=\frac{2*25}{5^2}=2[/tex]
m = 2 Kg
If the speed is doubled, v=10 m/s, the new kinetic energy is:
[tex]\displaystyle K=\frac{1}{2}2\cdot 10^2[/tex]
K = 100 J
If the speed is tripled, v=15 m/s, the new kinetic energy is:
[tex]\displaystyle K=\frac{1}{2}2\cdot 15^2[/tex]
K = 225 J
When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J
What is the velocity of v2 if m1=82kg v1= 25 m/s m2=60kg?
Given :
[tex]m_1 = 82 \ kg\\\\v_1 = 257 m/s\\\\m_2 = 60 kg[/tex]
Let, us assume that momentum is conserved during the process.
To Find :
The velocity [tex]v_2[/tex] .
Solution :
Applying conservation of momentum :
Initial momentum = Final momentum
[tex]m_1v_1 = m_2v_2\\\\82\times 257=60\times v_2\\\\v_2 = \dfrac{82\times 257}{60}\ m/s\\\\v_2 = 351.23\ m/s[/tex]
Hence, this is the required solution.