Answer:
Explanation:
The concept of a new strong nuclear force was introduced. In 1935, the first theory for this new force was developed by the Japanese physicist Hideki Yukawa, who suggested that the nucleons would exchange particles between each other and this mechanism would create the force.
please help me with this
Answer:
>400N is needed to balance that lever
please help me !! i’ll mark brainliest if you’re right!
What does it mean when a wave’s amplitude increases?
The wave’s wavelength gets longer.
The wave is moving through a denser medium.
The wave is carrying more energy.
The wave’s frequency also increases.
Answer:
the wave is carrying more energy
Explanation:
trust me broski
Answer:
a
Explanation:
Which type of border shows the division between Sonora and Chihuahua?
National border
State border
Physical boundary
Natural boundary
PLEASE HELP!!
5. A 700 kg race car travels
around the track at 65 m/s. The track has a radius of 75 m.
a. What is the centripetal acceleration?
b. What is the centripetal force?
c. Is the net force on the car less than, equal to, or greater than the centripetal force? Why?
Hi there!
a.
The equation for centripetal acceleration is as follows:
[tex]\large\boxed{a_c = \frac{v^2}{r}}}[/tex]
Plug in the given values to solve:
[tex]\large\boxed{a_c = \frac{(65)^2}{75} = 56.33 m/s^2}[/tex]
b.
According to Newton's Second Law:
[tex]\large\boxed{\Sigma F = ma}}[/tex]
The acceleration is v²/r, so the net force is:
[tex]\large\boxed{\Sigma F = m(\frac{v^2}{r})}}[/tex]
Multiply by the given mass:
[tex]\large\boxed{\Sigma F = 700(56.33) = 39433.33 N}}[/tex]
c.
There is NO net force in the vertical direction since the object is NOT accelerating in the vertical direction (normal force and weight cancel).
Thus, the ONLY net force experienced by the object is in the horizontal direction and is EQUAL to the centripetal force.
define nuclear energy
Answer:
Nuclear energy is the energy stored in atoms that can produce electricity.
Hope that helps. x
A car has a kinetic energy of 41.6 kJ.
The speed of the car is 8.0m/s.
Calculate the mass of the car.
A 7.5 kg beaver dives horizontally off a 50 kg log at a speed of 4 m/s. What is the speed of the log?
Answer:
[tex]0.6\; \rm m\cdot s^{-1}[/tex], assuming that drag of the water on the log is negligible.
Explanation:
The momentum [tex]p[/tex] of an object is equal to the product of mass [tex]m[/tex] and velocity [tex]v[/tex]. That is, [tex]p = m\, v[/tex].
If the drag of water on the log is negligible, momentum of the beaver and the log, combined, would be preserved.
The momentum of the beaver and the log, combined, was initially [tex]0\; \rm kg\cdot m \cdot s^{-1}[/tex].
The momentum of the beaver right after the dive would be [tex]7.5 \; {\rm kg} \times 4\; {\rm m \cdot s^{-1}} = 30\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
The sum of the momentum of the beaver and the log is conserved and should continue to be [tex]0\; {\rm kg\cdot m \cdot s^{-1}}[/tex] even after the dive. Since the momentum of the beaver is [tex]30\; {\rm kg \cdot m \cdot s^{-1}[/tex] after the dive, the momentum of the log should become [tex](-30)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Since the mass of the log is [tex]50\; {\rm kg}[/tex], the new velocity of this log would be:
[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{(-30)\; {\rm kg \cdot m \cdot s^{-1}}}{50\; {\rm kg}} \\ &= (-0.6)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
(The new velocity of the log is negative because the log would be moving away from the beaver.)
The speed of an object is the magnitude of velocity. For this log, a velocity of [tex](-0.6)\; {\rm m \cdot s^{-1}}[/tex] would correspond to a speed of [tex]|(-0.6)\; {\rm m \cdot s^{-1}| = 0.6\; {\rm m \cdot s^{-1}[/tex].
The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms (ng) in 1 g. What is the mass of this sample in ng? 2. 1 Times. 103 ng 2. 1 Times. 106 ng 2. 1 Times. 109 ng 2. 1 Times. 1012 ng.
2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.
Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.
There are other units of measure,
Milligram: 1 g is equal to the [tex]\bold {10^3 \ mg}[/tex]Micro-gram: 1 g is equal to [tex]\bold {10^{6} \ \mu g}[/tex]Nano-gram: 1 g is is equal to[tex]\bold {10^{9} \ ng}[/tex]First convert kg to gram,
Since, 1 Kg = 1000 g
2.1 kg = grams of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]
Now, convert 2100 g to nano-grams
Since, 1 g = 1 x 10⁹ ng
2100 g = ng of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]
Therefore, 2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.
To know more about Mass units,
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a 3 kg rock is falling from a rock ledge in the absence of air resistance how much force will the rock strikes the ground with
Answer:
papi sus
Explanation:
Which is the correct answer
Answer:
C
Explanation:
All of the other options are affecting the gravitational force and opposing force that pushed upwards, the diagram represents an object that is not moving up or down.
I desperately need help with my Physics Exam I am failing this class will mark Brainliest to whoever helps the most: PART 1 (I don't want to overwhelm anyone so I'm posting my questions in parts) Also the questions marked are incorrect, so just work with the 3 remaining answers.
10000000 thank you's
What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?
The half-life of the given isotope will be 10 days, if after 30 days only 31.25 grams are remaining from a sample of 250 grams of the sample size taken in the beginning.
What is Half-life of an element?
The Half-life is the time which is required for a quantity to reduce the content to half of the amount present as its initial value. The term is used in nuclear physics to describe how quickly an unstable atom undergo radioactive decay or how long does stable atoms survive. The term is also used generally to characterize any type of exponential decay.
The half-life of the isotope can be calculated by the formula:
FR = 0.5n
FR = Fraction Remaining = 31.25 g / 250.0 g = 0.1250
n = number of half lives elapsed = ?
0.125 = 0.5n
log 0.125 = n log 0.5
-0.9031 = -0.3010 n
n = 3.000 half lives have elapsed
3 half lives = 30 days
1 half live = 10 days
Therefore, the half-life of the isotope will be 10 days.
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A car travels 80 km at an avarage speed of 40 km/h. It travels the remaining distance in 3h. What is its dis placement if the average speed of the car is 30 km/h.? And explain
A) 100 km
B)120 km
C)150km
D)180km
When traveling with average speed 40 km/h, the car would cover a distance of 80 km in time t such that
40 km/h = (80 km) / t ⇒ t = 2 h
so the total travel time is 2 h + 3 h = 5 h.
Average speed is defined as the total distance traveled divided by the time it took to cover that distance. So if the overall average speed was 30 km/h, then
30 km/h = (80 km + x) / (5 h) ⇒ x = 70 km
where x is the distance traveled in the last 3 h of the trip.
Then the total displacement of the car is 80 km + 70 km = 150 km.
If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. maintain your speed and following distance speed up and decrease your following distance Submit answer
Answer:
The first one.
If the road becomes wet or crowded, you should ___ slow down and increase your following distance_
What is meant by surface distance ?
The safe distance between vehicles traveling in column specified by the command in light of safety requirements
The slower speed will help you save gas and avoid potential accidents. You can easily eliminate chances of rear end collision by maintaining a safe distance between your car and the vehicles ahead.
hence , a) slow down and increase your following distance is a correct option
learn more about safe distance
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The gravitational potential energy of a cucumber-harth system changes when which factor changes?
A) the cucumber's speed
B) the cucumber's mass
c) the cucumber's temperature
D) the length
in what direction will the seesaw rotate and what will the sign of the angular acceleration be?
Answer:
It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.
Starting from rest an object accelerates on a straight line at rate of 2ms‐² for 10 seconds. what is the speed of the object at the end of the 10 seconds
Answer:
20 meters/sec
Explanation:
See attachment.
2m/s^2 = (v - 0)/10s
v = 20 m/s
according to newton, doubling the distance between two interacting objects:
Answer:
So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).
Explanation:
hope his helps
Which of the following is an example of physical weathering?
O calcium carbonate in limestone changes to calcium hydrogen carbonate
O flow water carves erosion channels in a hillside
Ostalactites precipitate in a cave
O acid rain corrodes a monument
Answer:
acid rain corrodes a monument. d.
Explanation:it only makes sense because its doing something to a physical object.and the other ones aren't.
can somebody explain it to me please?
Range be R and height be h
[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]
[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]
u=initial velocity
theta is angle of projection.
g=acceleration due to gravity
ATQ
[tex]\\ \sf\longmapsto R=2h[/tex]
[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]
Cancelling required ones
[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]
sin2O=2sinOcosO
[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]
[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]
[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]
[tex]\\ \sf\longmapsto tan\theta=2[/tex]
[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]
[tex]\\ \sf\longmapsto \theta=63.4°[/tex]
Done
Option B is correct
A race car makes one lap around a track of radius 180 m in 90.0 s. What is the centripetal acceleration in m/s2?
a
0.5
12.6
Ob
OC
Od
.87
2.
2
Answer:
0.9m/s^2 (yours is 0.87, so choose that)
Explanation:
formula for centripetal acceleration:
v^2/r
to find v, we know that f=1/90s, and r=180m.
v=(2pir)/T
v=(2pi(180))/90
v=12.6m/s
now plug into a=v^2/r
a=(12.6)^2/180
a=0.9m/s^2
how much force would be needed to push a box weighing 30 N up a ramp that ahas an ideal mechanical advantage of 3
Answer:
60n would be the needed force
The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.
What is the mechanical advantage?The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.
The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.
The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.
Given the load = 30 N and the ideal mechanical advantage = 3
Mechanical advantage = Load/ Effort
Input force or effort = Load/ M.A.
Force = 30/3
Input Force = 10 N
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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____?
[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]
Projectile is thrown with a velocity = v Angle of projection = θVelocity of projectile at a height half of the maximum height covered be [tex] \sf{v_0}[/tex][tex]\qquad[/tex]______________________________
Then –
[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]
Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]
Therefore, the vertical component of velocity of projectile at this height will be–
☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]
Answer:
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2
Which would ba another example of newtons first law?
can someone explain it with steps?
A car was moving on a road at a constant speed of 15 m/s when suddenly the car driver saw some animal on the road at a distance of 21 m from the car, so he applied the brakes after a response time of 0.4 s and stopped before hitting the animal by 1 m. What was the deceleration of the car?
a-7.5 m/s^2
b-5.2 m/s^2
c-8.0 m/s^2
d-5.6 m/s^2
Answer:
Option C is the correct answer
Explanation:
Distance travelled by car during reaction time
[tex]=15\times0.4\\\\=6m[/tex]
The car stopped before hitting the animal by [tex]1 m[/tex]
Distance travelled during deceleration is [tex]21-6-1=14m[/tex]
Hence by [tex]v^2=u^2+2as[/tex]
We have
[tex]0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2[/tex]
Option C is the correct answer
Distance traveled during reaction time
15(0.4)=6mTotal distance
21-6-1=14m[tex]\\ \sf\longmapsto v^2-u^2=2as[/tex]
[tex]\\ \sf\longmapsto -(15)^2=2(14)a[/tex]
[tex]\\ \sf\longmapsto -225=28a[/tex]
[tex]\\ \sf\longmapsto a=-8.0m/s^2[/tex]
Why is kinetic energy lost in an inelastic collision?
Answer:
This is because some kinetic energy had been transferred to something else.
Explanation:
An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not.
Definition of main energy level
Answer:
the orbital in which the electron is located relative to the atom's nucleus.
Explanation:
Answer:
the principal energy level of an electron refers to the shell or orbital in which the electron is located relative to the atom's nucleus.
What is number 10 for this?
Answer:
The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.
Given data in the question;
Speed of wave;
Frequency of wave;
wavelength;
To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.
Where is wavelength, f is frequency and c is the speed.
We substitute our given values into the equation
Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.
Explanation:
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2. 15, 2. 05, 02. 02, 02. 7. Use the table to answer the questions. What is the fastest time trial for the first quarter checkpoint? seconds What is the slowest time trial for the first quarter checkpoint? seconds What is the range of times measured for this checkpoint? seconds.
The fastest time trial for the first quarter checkpoint is 2.02 s.
The slowest time trial for the first quarter checkpoint is 2.7 s.
The range of the times measured for the checkpoint is 0.68 s.
The given parameters;
Time for quarter checkpoint, = 2.15, 2.05, 2.02, 2.7The fastest time trial for the first quarter checkpoint is the least measured time value.
fastest time trial = least time measured
fastest time trial = 2.02 s
The slowest time trial for the first quarter checkpoint is the highest measured time value.
slowest time trial = 2.7 s
The range of the times measured for the checkpoint is difference between the fastest time and slowest time.
Range = fastest time - slowest time
Range = 2.7 - 2.02
Range = 0.68 s
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Answer:
2.02, 2.15, 0.13
Explanation:
A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.
Use the table to answer the questions.
What is the fastest time trial for the first quarter checkpoint?
2.02
seconds
What is the slowest time trial for the first quarter checkpoint?
2.15 seconds
What is the range of times measured for this checkpoint?
0.13 seconds