Answer:
A = 2 m from fulcrum
Explanation:
Product of anti clockwise = Product of clockwise moment
5 × 4 = 10 × A
20 = 10 x A
A = 20 / 10
A = 2 m from fulcrum
A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note: Csilver = 234J/kg.°CL
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
1. A boy is standing on top of a roof that is 4.5 m high. If he throws a 1.4 kg water balloon down with a velocity of 2m/s,
what velocity will the water balloon strike the ground with? (ignore air resistance)
Answer:
The magnitude of the final velocity of the water balloon is 84.3 m/s.
Explanation:
To calculate the velocity of the water balloon when it strikes the ground we need to use the next equation:
[tex] v_{f}^{2} = v_{0}^{2} - 2gh [/tex]
Where:
[tex]v_{f}[/tex]: is the final velocity =?
[tex]v_{0}[/tex]: is the initial velocity = 2 m/s
g: is the gravity = 9.81 m/s²
h: is the height = 4.5 m
By entering the above values into equation (1) we have:
[tex] v_{f}^{2} = v_{0}^{2} - 2gh = (2 m/s)^{2} - 2*9.81 m/s^{2}*4.5 m = -84.3 m/s [/tex]
The minus sign is because the direction of the velocity vector is in the negative vertical axis.
Therefore, the magnitude of the final velocity of the water balloon is 84.3 m/s.
I hope it helps you!
The object has a speed of 15 m/s. It took the object 5 second to travel, how far did the object go?
Answer:
Seventy-Five Metres
Explanation:
You can solve this simply by multiplying 15 m/s by 5 seconds
[tex]15 \frac{m}{s} * 5s\\=75\frac{ms}{s}\\= 75m[/tex]
A whale comes to the surface to breathe and then dives at an angle 24 degrees to the horizontal surface of the water. The whale continues in a straight line 145 m. What are the horizontal and vertical components of the displacement of the whale?
Given that,
A whale dives at an angle of 24 degrees to the horizontal surface of the water.
The whale continues in a straight line 145 m.
To find,
The horizontal and vertical components of the displacement of the whale.
Solution,
The horizontal component of displacement is :
[tex]d_x=d\cos\theta\\\\=145\times \cos(24)\\\\=132.46\ m[/tex]
The vertical component of displacement is :
[tex]d_y=d\sin\theta\\\\=145\times \sin(24)\\\\=58.97\ m[/tex]
Hence, the horizontal and vertical components of the displacement of the whale are 132.46 m and 58.97 m.A 100 A current circulates around a 2.00-mm-diameter superconducting ring.
A. What is the ring's magnetic dipole moment?
B. What is the on-axis magnetic field strength 4.70 cm from the ring?
Answer:
0.000314 Am²
6.049*10^-7 T
Explanation:
A
From the definitions of magnetic dipole moment, we can establish that
= , where
= the magnetic dipole moment in itself
= Current, 100 A
= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²
On solving, we have
= ,
= 100 * 3.14 * 0.000001
= 0.000314 Am²
B
= (0)/4 * 2 / ³, where
(0) = constant of permeability = 1.256*10^-6
z = 4.7 cm = 0.047 m
B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]
B = 1*10^-7 * 0.000628/1.038*10^-4
B = 1*10^-7 * 6.049
B = 6.049*10^-7 T
an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any vertical deviations the hammer moves at the rate of 1.16 rev/s what is the tension in the chain
Answer:
T = 692.42 N
Explanation:
Given that,
Mass of hammer, m = 8.71 kg
Length of the chain to which an athlete whirls the hammer, r = 1.5 m
The angular sped of the hammer, [tex]\omega=1.16\ rev/s=7.28\ rad/s[/tex]
We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :
[tex]F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N[/tex]
So, the tension in the chain is 692.42 N.
Answer5 ms 3. A football player has a mass of 95 kg, and he is running with a velocity of 15 m/s. What is his momentum? Answer:
Answer:
1425kgm/s
Explanation:
Given parameters:
Mass = 95kg
Velocity = 15m/s
Unknown:
Momentum = ?
Solution:
The momentum of a body is the amount of motion it posses;
Momentum = mass x velocity
Insert the parameters and solve;
Momentum = 95 x 15 = 1425kgm/s
If a ball leaves the ground with a velocity of 4.67 m/s,
how high does the ball travel?
Answer:
[tex]Vf^2=Vo^2+2aS\\(0m/s)^2=(4.67m/s)^2+(2*-10m/s^2)S\\-(4.67)^2 m^2/s^2=-20m/s^2*S\\S=(21.8089/20) m\\S=1.090445 m\\[/tex]
19 "Made mostly of ice", fits best under which heading in the table below?
Asteroid
Meteor
Comet
Planet
Description
A Asteroid
B Meteor
C Comet
D Planet
In a tug of war, team A pulls the rope with a force of 50 N to the right and team B pulls it with a force of 110 N to the left. What is the resultant force on the rope and which direction does the rope move? (N represents newton, the unit of force)
Given that,
Force in right side = 50 N
Force in left side = 110 N
To find,
The resultant and direction of force.
Solution,
Let the net force F is acting on the rope. Also, we can assume that right side is positive and left side is negative.
F = 50+(-110)
= -60 N
So, the magnitude of the resultant force acting on the rope is 60 N and it acts in left side.
Two forces of 50 N and 30N respectively, are acting on an object. Find the net force (in N) on the object if
A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions
Answer:
A) 80 N
B) 20 N
Explanation:
A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..
Thus;
ΣF = 50 + 30
ΣF = 80 N
B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;
ΣF = 50 - 30
ΣF = 20 N
Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a resting point midway between the two particles, an external agent must do work equal to
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
A horizontal wire PQ is perpendicular to a uniform horizontal magnetic field. A length of 0.25 m of the wire is subject to a magnetic field strength of 40 mT. A downward magnetic force of 60 mN acts on the wire. What is the magnitude and direction of the current in the wire
Complete question:
Please find the image uploaded for the diagram.
Answer:
The magnitude of the current is 6 A, in the direction of Q to P.
Explanation:
Given;
length of the wire, l = 0.25 m
magnetic field strength, B = 40 mT = 0.04 T
Magnitude of the magnetic force, F = 60 mN = 0.06 N
The magnitude of the current in the conductor is given as;
F = BIL
Where;
I is the current induced in the conductor
I = F / BL
I = (0.06) / (0.04 x 0.25)
I = 6 A
Th current will move in direction of Q to P
A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
Answer:
132 N
Explanation:
Given that a 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
From Newton 2nd law of motion,
Change in momentum = impulse.
Change in momentum = m( V - U )
Substitute all the parameters into the formula
Change in momentum = 1.1 ( 4.5 - 1.5 )
Change in momentum = 1.1 × 3
Change in momentum = 3.3 kgm/s
Impulse = Ft
That is,
Ft = 3.3
Substitute time t into the formula above
F × 0.025 = 3.3
F = 3.3 / 0.025
F = 132 N
Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.
Help Me Please
8th grade science, one question
A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How far did the bullet travel horizontally when it hit the ground?
595.2 m
478.4 m
364.0 m
247.2 m
Answer:
im pretty sure its a
Explanation:
In a car, how does an air bag minimize the force acting on a person during a collision?
Answer:
It increases the time it takes for the person to stop.
Explanation:
Answer:
C: It increases the time it takes for the person to stop.
Explanation:
on edge! hope this helps!!~ o(〃＾▽＾〃)o
why the efficiency of the single movable pully system is not 100%
Answer:
This is because part of the energy input is used to overcome gravity, inertia and friction
A marble is fired from a spring-loaded gun at an angle of 30◦
above the ground and reaches a
maximum height 500m.
a. What is the velocity of the marble at it’s maximum height?
b. What is the acceleration of the marble at it’s maximum height?
c. What was the marbles initial velocity?
d. How long did it take for the marble to reach it’s maximum height?
Answer:
jrirnr tn
Explanation:
and then we was just wondering what we should be talking to about to see how you are
Two train whistles have identical frequencies of 220 Hz. When one train is at rest in the station sounding its whistle, a beat frequency of 10.0 Hz is heard from the other train that is approaching the station. What is the speed of the approaching train? Assume that the speed of sound is 340 m/s.
Answer:
Speed of the approaching train = 15.45 m/s
Explanation:
Given:
Frequency F0 = 220 Hz
Beat frequency F1 = 10.0 Hz
Find:
Speed of the approaching train
Computation:
Approaching frequency F2 = 220 + 10.0 Hz
Approaching frequency F2 = 230 Hz
Doppler shift;
F = [(v+v0)/(v-vS)]F0
230 = [(340+v0)/(340-0)]220
V0 = 15.45 m/s
Speed of the approaching train = 15.45 m/s
A crate of oranges on a horizontal floor has a mass of 30 kg. The coefficient of static friction is 0.62. The coefficient of kinetic friction is 0.52. The worker pulls the crate with a force of 200 N.
How do you know if the crate is moving or not? Explain it by comparing the magnitude of the friction with the magnitude of the applied force.
Answer:
Explanation:
The Net Force
The net force is defined as the sum of all the forces acting on a body at a certain moment.
Recall some basic concepts and equations to solve the problem.
If no external forces are applied in the vertical direction, the weight of the object and the normal force have the same magnitude and point to opposite directions.
The friction force is defined as:
[tex]Fr_k=\mu_k N[/tex]
[tex]Fr_s=\mu_s N[/tex]
Where the subindices k and s are referred as to the kinetic and static friction forces respectively.
The condition for the object to move is that the applied force is greater than the friction force.
The crate of oranges has a mass of 30 Kg, thus its weight is:
W = m.g = 30 * 9.8 = 294 N
The normal force is:
N = W = 294 N
The kinetic friction is calculated as:
[tex]Fr_k=0.52* 294[/tex]
[tex]Fr_k=152.88\ N[/tex]
The static friction is calculated as:
[tex]Fr_s=0.62* 294[/tex]
[tex]Fr_s=182.28\ N[/tex]
The applied force is 200 N and it's greater than both the kinetic and the static friction forces, thus the crate is definitely moving at a certain positive acceleration.
If the lead bar was heated to a temperature of 120 degrees Celsius, what would happen to its volume, mass and density?
On heating a lead bar, its volume increases, density decreases and its mass remains constant.
According to Ideal gas equation,
pV = nRT
ρ = m / V
V = Volume
T = Temperature
ρ = Density
m = Mass
Based on Ideal gas equation, Volume is directly proportional to Temperature. As temperature increases, volume also increases. Based on density formula, Volume is indirectly proportional to Density. As Volume increases density decreases.
In any reaction, the mass is always conserved. Because, mass is the amount of matter present in a substance. During a reaction, atoms are neither created nor destroyed, just rearranged. So the mass always remains constant.
Therefore, if a lead bar was heated to 120°C, its volume increases, density decreases and its mass remains constant.
To know more about Ideal gas equation
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0
Northeast
North Central
South
Region of the United States
West
●
8 Claims Evidence Reasoning Which
region of the country has the coldest
January temperatures? Use the data in the
graph to support your claim, and explain
A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 7.1 m making 3.9 revolutions every 9.4 second. What is the centripetal acceleration of the ball?
Answer:
48,2 m/s²
Explanation:
We're gonna use the Centripetal Acceleration formula: v² / r but before that, we got to know the velocity, that is not shown clearly to us, so....
To know the velocity let's calculate the distance that the ball traveled
The circumference of a circle formula is:
2piR
2 . 3,14 . 7,1 | That is equal to 44,588 m
We know that the ball traveled this distance 3,9 times, so...
44,588 . 3,9 = 173,8932 m
Ok, now we have the distance, just need to know the time, that is 9.4 seconds.
Velocity = Distance / Time
V = 173,8932 / 9,4
V = 18,5 (approximate)
So...
We are back to the first formula:
Ca = v² / r
Ca = 18,5² / 7.1
Ca = 48,2 m/s² (approximate)
I hope it is correct, hahaha.
Energy stored because of an object's height above the Earth's surface is_____energy.
nuclear
gravitational
electrical or chemical
A force of 3.40 N is exerted on a 6.30 g rifle bullet. What is the bullet's acceleration?
The correct answer is a = 539.68 m/[tex]s^2[/tex]
A push or pull that one thing applies to another is known as force. An object is considered to exert a force when it accelerates through space because acceleration is the rate at which an item's speed changes. The second law of motion of Newton explains this concept.
The equation F=ma, where "F" stands for force, "m" for mass, and "a" for acceleration, illustrates the link between force and acceleration.
Force applied = F = 3.40 N
The bullet's weight = m = 6.30 g = 0.0063 kg
The bullet's rate of acceleration can be expressed as = a = F/m
a = 3.40 / 0.0063
a = 539.68 m/[tex]s^2[/tex]
To learn more about acceleration refer the link:
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Thomas the Tank Engine (a train) is going 80 m/s and slows down to 30 m/s over a period of 30s. What is his deceleration? Acceleration= (final velocity-initial velocity)/ time A. -1.67 m/s/s B. 0.67 m/s/s C. -50 m/s/s D. 50 m/s/s
Answer: D
Explanation:
Can someone please help me with some physics
Answer:
sure I will helpy you iru
Carlos is riding her bicycle. His velocity is 5 m/s to the right. His acceleration is 1.2 m/s2 to the left. He is doing what?
- moving at a constant speed
- slowing down
- speeding up
Answer:
Slowing down
Explanation:
Carlos is travelling to the right as indicated by his velocity, so his velocity is +5m/s. He is also accelerating to the left which is the opposite direction to his velocity so his acceleration is -1.2m/s2 to the right.
Since his acceleration is negative relative to his velocity and direction of motion, his speed is decreasing and he is slowing down.
Hope this helped!
A cannon can make a maximum angle of 30 degrees with the horizon. What is the minimum speed of a cannon ball if it must clear a 10-m-high obstacle 30 m away?
1. 28.3 m/s
2. 30.5 m/s
3. 32.8 m/s
4. 34.1 m/s
5. 36.8 m/s
Answer:
28.3 m/s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 30°
Maximum height (H) = 10 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
Thus, we can obtain the minimum velocity cannon ball by using the following formula:
H = u²Sine² θ / 2g
10 = u² × (Sine 30)² / 2× 10
10 = u² × (0.5)² / 20
10 = u² × 0.25 / 20
10 = u² × 0.0125
Divide both side by 0.0125
u² = 10/ 0.0125
u² = 800
Take the square root of both side
u = √800
u = 28.3 m/s
Therefore, the minimum speed of the cannon ball is 28.3 m/s