Answer:
20m/s
Explanation:
u = 0m/s
t = 2s
a = +g = 10m/s²
t = 2d
v = ?
v = ut + 1/2at²
v = 0(2) + 1/2(10)(2)²
v = 0 + 5(4)
v = 20m/s
20 POINTS
In order to maximize the acceleration of an object, one should
maximize the mass
maximize the force
minimize the velocity
maximize the inertia
What
A moving object always has energy in its
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is present, given by B = 4.0i + 9.0x2 j , with x in meters and B in mT. Calculate the k-component of the force on the 2 m segment of the conductor that lies between x = 1.0 m and x = 3.0 m.
Answer:
0.546 [tex]\hat k[/tex]
Explanation:
From the given information:
The force on a given current-carrying conductor is:
[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]
where the length usually in negative (x) direction can be computed as
[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]
[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]
[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]
where;
current I = 7.0 A
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]
F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]
F = 0.546 [tex]\hat k[/tex]
What is exerted when atoms collide with the walls of their container?
O gravity
O pressure
O stress
O friction
Answer:
B
Explanation:
Answer:
Pressure
Explanation: I took a test
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. What is the vertical component of the ball’s velocity? What is the horizontal component of the ball’s velocity?
0 m/s; 70 m/s
61.8 m/s; 32.9 m/s
32.9 m/s; 61.8 m/s
70 m/s; 0 m/s
Answer:
answer is 61.8 m/s; 32.9
l am not sure
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, answer 1 is the correct answer.
What happens to the electrical energy that does not become light energy?
A. The lightbulb transforms it into mechanical energy,
B. The lightbulb transforms it into thermal energy.
C. Some of the energy is destroyed over time rather than being
conserved
D. New energy is produced in the system when the lightbulb creates
light energy
- PREVIOUS
NEXT->
Answer:
B
Explanation:
To solve this we must be knowing each and every concept related to electrical energy. Therefore, the correct option is option B among all the given options.
What is electrical energy?The work done by an electric charge is referred to as electrical energy. For time t seconds, if electricity I ampere passes throughout a conductor or any other conducting element with a potential differential v volts across it.
The kilowatt hour is both a practical as well as an economic unit of electrical energy. The basic commercial unit is the watt-hour, one and kilowatt hour equals 1000 watt hours. Electric supply providers charge their customers every kilowatt hour unit of electricity used. This kilowatt hour seems to be a BOT unit, or board of trade unit. The electrical energy that does not become light energy, the lightbulb transforms it into thermal energy.
Therefore, the correct option is option B among all the given options.
To know more about electrical energy, here:
https://brainly.com/question/1698090
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A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1 .What is the average acceleration of the ball?
Answer:
50m/s^2Explanation:
Step one:
given data
initial velocity u= 0m/s since the ball is at rest
time of contact t= 0.3s
final velocity v=15m/s
Required
acceleration a
from the first law of motion
v=u+at
substitute our given data
15=0+a*0.3
15=0.3a
divide both sides by 0.3
a=15/0.3
a=50m/s
The average acceleration is 50m/s^2
True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
which thermometer is used in hot region.why?
Answer:
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.
Explanation:
please mark me brainlist
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.
My favorite Naruto couple fighting together!!!! Remember this?! And remember.........this?!
I cried so hard when he died.
Answer:
ok tbh i didnt cry when neji died but now when i think about it i just cry thinking about how himawari could have met her uncle
Explanation:
what does the modal "must"indicate?
Answer:
The modal verb must is used to express obligation and necessity. The phrase have to doesn't look like a modal verb, but it performs the same function.
Please help
A golfer hits a ball is at 15 m/s at an angle of 40 degrees above the horizontal. How far from where the ball was
hit will the ball land?
Answer:
Range = 22.61 m
Explanation:
We can use the formula for the Range in flat ground, given by:
[tex]Range=v_i^2\frac{sin(2\theta)}{g}[/tex]
which for our case renders:
[tex]Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m[/tex]
By using the range formula in the motion of a projectile, the ball will land 22.6 meters from where it was hit.
Given that a golfer hits a ball at 15 m/s at an angle of 40 degrees above the horizontal, To calculate how far the ball travelled, we will use the range formula to calculate the total distance travelled.
Range R = [tex]u^{2}[/tex]sin2∅ /g
Where
u = 15 m/s
g = 9.8 m/[tex]s^{2}[/tex]
∅ = 40 degrees
Substitute all the parameters into the above formula.
R = [tex]15^{2}[/tex]sin(2 x 40) / 9.8
R = 225sin80/9.8
R = 221.58/9.8
R = 22.6 m
Therefore, the ball will land 22.6 meters from where it was hit.
Learn more here: https://brainly.com/question/25299583
What voltage is required to move 6A through 5Ω?
A locomotive creates a
59,400 N force, which creates
an acceleration of 0.145 m/s.
What is the mass of the
locomotive?
Unit=kg
Answer:
410,000 kg
Explanation:
Use Newton's second law
F = ma
m = F/a
m = 59,400 N/(.145 m/s) = 410,000 kg
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
Answer:
a
[tex]\lambda = 3.68 *10^{-36} \ m[/tex]
b
[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 180 \ kg[/tex]
The speed of the person is [tex]v = 1 \ m/s[/tex]
The energy of the proton is [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]
Generally the de Broglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m * v }[/tex]
Here h is the Planck constant with the value
[tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]
=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]
Generally the energy of the proton is mathematically represented as
[tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]
Here [tex]m_p[/tex] is the mass of proton with value [tex]m_p = 1.67 *10^{-27} \ kg[/tex]
=> [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]
=> [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]
=> [tex]v = 3.09529 *10^{7} \ m/s[/tex]
So
[tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]
so [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]
=> [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it around her head in a horizontal circle at 100 rpm. Assume the room temperature is 20 degrees Celsius. What are the highest and lowest frequencies heard by a student in the classroom?
Answer:Highest frequency =618.89Hz
Lowest frequency=582.22Hz
Explanation:
The linear velocity of a sound generator is related to angular velocity and is given as
Vs = rω where
r = the radius of circular path = 1.0 m
ω is the angular velocity of the sound generator. = 100 rpm
1 rev/min = 0.10472 rad/s
100rpm =10.472 rad/ s
Vs = rω
= 1m x 10.472rad/ s= 10.472m/s
A) Highest frequency heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,
f max = (v/ v-vs) fs
Where , v is the speed of the sound in air at 20 degrees celcius =
343 metres per second
vs is the linear velocity of the sound generator=10.472m/s
fs is the frequency of the sound generator= 600 Hz
f max = (343/ 343 - 10.472) x 600
=343/332.528) x600
=618.89Hz
B) Lowest frequency heard by a student in the classroom = Minimum frequency
f min = (v/ v+vs) fs
(343/ 343 + 10.472) x 600
=343/353.472) x 600
=582.22hz
A 50 kg person steps off a diving platform that is 10 meters above the water below (Olympic height). With what speed do they hit the water?
Answer:
14 m/s
Explanation:
The following data were obtained from the question:
Mass = 50 kg
Initial velocity (u) = 0 m/s
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
The velocity (v) with which the person hit the water can be obtained as shown below:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 10)
v² = 0 + 196
v² = 196
Take the square root of both side
v = √196
v = 14 m/s
Therefore, he will hit the water with a speed of 14 m/s
Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?
Answer:
(a) the change in length of the silk is 0.001585 cm
(b) the maximum weight that a single thread can support is 17.67 N
Explanation:
Given;
mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg
length of the silk, L = 12 mm = 0.012 m
diameter of the silk, d = 0.15 mm
radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m
The cross sectional area of the silk;
A = πr² = π(0.075 x 10⁻³)²
A = 1.767 x 10⁻⁸ m²
The Young's modulus of elasticity of spider-silk is given by;
2.1 Gpa = 2.1 x 10⁹ N/m²
(a)
Apply Young's modulus of elasticity equation to determine the change in length of the silk;
[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]
[tex]x = 0.001585 \ cm[/tex]
(b)
the maximum weight that a single thread can support is given by;
[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]
The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²
[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]
A car accelerates steadily from 15 m/s to 25 m/s in 5 s. what us its speed
A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?
A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s
Answer:
B. 1275 kg*m/s
Explanation:
I = F(deltaT) = (deltaP) = mv2- mv1
Therefore,
I = mv2-mv1
m = 850 kg
v2 = 5 m/s
v1 = 3.5 m/s
I = (850)(5)-(850)(3.5)
I = 1275 kg* m/s
Please help!How is constant or uniform acceleration used to explain free fall?
How can you find directions using satellite orbiting?
Answer:
Satellites may move north to south, or south to north, or west to east, but never from east to west. When satellites are launched, they always head eastward to take advantage of the Earth's rotation, going more than 1,000 miles per hour near the equator. This saves a lot of fuel.
10points asap
A force of 30 N acts upon a 7 kg block. Calculate its acceleration.
Protists are unique organisms that are so different from each other that they are sometimes called the 'junk drawer' kingdom.
True or False
Answer:
true
Explanation:
Im in k12 and I got an 100%
The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
Using the dimensions;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s
A marble is rolling off the edge of a table is observed to hit the floor 0.77 m from the table. If the top is 0.86 m high, how fast was the marble traveling when it left the table?
A.) 0.2 m/s
B.) 0.4 m/s
C.) 1.8 m/s
D.) 1.6 m/s
Answer:
B 0.4 have a nice day and hiiii
If the weight of a person is 500 newton what is his mass on the earth ?
Answer:
The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?
f= Force (of gravity)=500N
g=acceleration of gravity=1.6m/s^2
m=mass=312kg
m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg
his mass is 312kg
A spring-loaded toy gun is used to shoot a ball straight up inthe air. The ball reaches a maximum height H, measuredfrom the equilibrium position of the spring. The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time?
Answer:
H' = H/4
Explanation:
By applying the law of conservation of energy to this problem, we know that:
Elastic Potential Energy Stored by Spring = Gravitational Potential Energy of ball
(1/2)kx² = mgH
H = (1/2)kx²/mg -------------- equation (1)
where,
H = Height reached by the ball
x = compression of spring
k = stiffness of spring
m = mass of ball
g = acceleration due to gravity
Now, if we make the compression to half of its value:
x' = x/2
then:
H' = (1/2)k(x/2)²/mg
H' = (1/4)(1/2)kx²/mg
using equation (1), we get:
H' = H/4