The others are:
Ampere
Mol
Candela
Answer:
Ampare mole candelahope it is helpful to you
The sodium-potassium pump moves blank ions into the cell and blank ions out of the cell.
potassium ions into the cell and
outside
⇒ sodium ions out of the cell.
Select all the correct answers.
Which situations describe an elastic collision?
(A) Two glass marbles bounce off each other.
(B) Rodrick flops onto his sofa and sinks into the cushion.
(C) A tossed water balloon flattens when it lands on the grass.
D) A bowling ball knocks over five pins.
I need help I do test I scared
Answer:
i think the answer is #3. :)
Why does it make sense that a 9 Liter pot of boiling water has more heat than a 1.5 L pot of boiling water?
Explanation:
because 9 is more than 1.5 and the more water, the more heat
What is the difference
between interplanting and intercropping?
Answer:
Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.
Intercropping refers to large, often commercial, farming operations and is the procedure of planting alternating crop rows. Growing two or more crops in close proximity is known as intercropping.
What is interplanting?Starting to grow two or more crops in close vicinity is known as intercropping.
The most common goal of intercropping is to increase yield on a given plot of land by utilizing resources that would otherwise go unused by a single crop.
Intercropping is the cultivation of more than one crop in the same space at the same time. Intercropping is the simultaneous planting of two or more species in a mixture, or the interplanting of one species during the growth of another.
The terms are frequently and regionally used interchangeably. Intercropping is the practice of planting alternating crop rows in large, often advertising, farming operations.
Thus, this is the difference between intercropping and interplanting.
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Christina drives her moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East.
What distance did she cover? Please include the number and unit.
What was the magnitude of her displacement? Please include the number and unit.
What was the direction of her displacement? Please abbrieviate (N, S, E, W, etc.)
Explanation:
Given that,
Christina drives her moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. At first she traveled North and then East. They are perpendicular to each other.
Total path traveled = distance covered by an object
D = 7 km + 5 km = 12 km
Shortest path covered = Displacement
[tex]d=\sqrt{7^2+5^2} \\\\d=8.6\ km[/tex]
For direction :
[tex]\tan\theta=\dfrac{d_y}{d_x}\\\\\tan\theta=\dfrac{7}{5}\\\\\theta=\tan^{-1}(\dfrac{7}{5})\\\\\theta=54.46^{\circ}[/tex]
She is moving in the direction of 54.46 degrees.
I will mark brainliest !!
Answer:
i think its b
Explanation:
A 4 kg jug of juice is pulled forward across a horizontal breakfast table, by a 12 N forward-directed force. A frictional force also impedes the jug's motion. What is the coefficient of kinetic friction between the table and jug, if the jug accelerates to the right at 2 m/s^2 ?
A 0.08
B 0.1
C 0.2
D 0.3
Answer:
0.1Explanation:
According to equation of motion;
\sum F = ma
m is the mass of the body
a is the acceleration
\sum F = Fm - Ff (along the horizontal)
Fm is the moving force
Ff is the frictional force
Given;
Fm = 12N
Ff = nR = nmg
Ff = n(4)(9.8)
Ff = 39.2n
Substitute the given value into the expression above;
Fm - Ff = ma
12 - 39.2n = 4(2)
-39.2n = 8-12
-39.2n = -4
n = 4/39.2
n = 0.1
Hence the coefficient of kinetic friction between the table and jug is 0.1
An airplane flies north at 300.0 km/h relative to the air and the wind is blowing south at 15.0 km/h. What is the airplane’s velocity relative to the ground? *
315 km/h North
15 km/h south
285 km/h North
300 km/h South
Answer:
285
Explanation:
read and alot jbgdxtffhgdae
F = M x G
Find the force of gravity acting upon a 1500Kg Hippopatamus.
Answer:........... .. .....
Explain why clear-cutting is a more destructive method of wood harvesting than selective cutting.
Answer:
Clear-cutting eliminates all trees leaving none behind to maintain soil fertility or house native animals.
Explanation:
edge 2022
A forestry/logging practice known as clear-cutting, clear felling, or clear-cut logging involves consistently cutting down the majority or all of the trees in a given region. Foresters utilize it to cultivate particular types of forest ecosystems and to advance particular species, in addition to harvesting shelter wood and seed trees.
What is deforestation?The removal of a forest or tree stands from land before it is used for a purpose other than as a forest is known as deforestation or clearing of forests. The use of forest land for urban, agricultural, or ranch purposes is referred to as deforestation. In tropical rainforests, there is the greatest concentration of deforestation. Forests currently cover about 31% of the Earth's land area. With half of that loss occurring in the previous century, this is a third less forest cover than there had before agriculture's growth.
As a result of clear-cutting, no trees are left to protect the topsoil or serve as a haven for wildlife. In turn, this causes soil erosion and soil fertility to be destroyed, as well as the removal of the trees required to produce seeds for natural forest development to continue. Animal populations are also pushed out of their natural habitats by clear-cutting, forcing them to look for new homes.
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which two technologies use waves to determine the location and velocity of objects
Lenses
X Rays
Sonar
Radar
Answer: sonar and radar
Explanation:
Picture
What is Humpty Dumpty's momentum if he has a mass of 56kg and is walking to the wall at a rate of 3 m/s?
Answer:
168
Explanation:
56 x 3
have a nice day
If white light is a mixture of all the primary colours of light, what is black?
Answer:
The primary colors of light are red, green, and blue. If you subtract these from white you get cyan, magenta, and yellow. ... Mixing these three primary colors generates black. As you mix colors, they tend to get darker, ending up as black.
A stone falls freely from rest for 8.0s what is it final velocity
can somone pls help me??!! i’m very stuck
Answer:
b
Explanation:
A 5 kg ball is sitting on the floor without moving. What is the momentum of the ball?
Answer:
The ball has no momentum
Explanation:
The given parameters are;
The mass of the ball = 5 kg
The velocity of the ball = 0 (The ball is sitting on the floor without moving)
The momentum of the ball = The mass of the ball × Velocity of the ball
Therefore, the momentum of the ball = 5 kg × 0 m/s = 0
The momentum of the ball is zero, the ball has no momentum.
An airplane is flying with velocity of 70km\hr in north east direction .The wind is blowing 30km\hr from north to south.What is the resultant displacement of the aeroplane in 4 h
Answer:
The resultant displacement of the airplane in 4 hours is 212.8 km.
Explanation:
The components of the airplane's velocity and wind's velocity are:
Airplane:
[tex] v_{a_{x}} = v_{a}cos(45) = 70 km/h*cos(45) = 49.50 km/h [/tex]
[tex] v_{a_{y}} = v_{a}sin(45) = 70 km/hsin(45) = 49.50 km/h [/tex]
Wind:
[tex] v_{w_{x}} = 0 [/tex]
[tex] v_{w_{y}} = v_{w} = -30 km/h [/tex]
Now, to know the new velocity of the airplane we to find the result vector:
[tex] v_{x} = v_{a_{x}} + v_{w_{x}} = 49.50 km/h + 0 = 49.50 km/h [/tex]
[tex]v_{y} = v_{a_{y}} + v_{w_{y}} = 49.50 km/h - 30 km/h = 19.50 km/h[/tex]
Now, the magnitude of the new speed of the airplane is:
[tex] v_{a} = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(49.50 km/h)^{2} + (19.50 km/h)^{2}} = 53.20 km/h [/tex]
Finally, after 4 hours the resultant displacement of the airplane is:
[tex] x = v*t = 53.20 km/h*4 h = 212.8 km [/tex]
Therefore, the resultant displacement of the airplane in 4 hours is 212.8 km.
I hope it helps you!
An objec with a mass of 7 kg accelerates 5m/s2 when an unknown force is applied to it.what is the magnitude of the unknown force?
Answer:
35 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 7 × 5
We have the final answer as
35 NHope this helps you
a 5 kg mass is accelerated at a rate of 2.5 m/s/s. what is the force that is exerted on this object
Answer:
12.5 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 5 × 2.5
We have the final answer as
12.5 NHope this helps you
A beam of electrons moving with a velocity Of 101 m/s
enters midway between two horizontal parallel plates which
Que 5cm long. 2cm apart and have pod Of V volt
applied between them - Calculate v if the beam is deflected
so that it just grazes the edge of the lower plate.(e/m = 1.8×10^11 c/kg)
Answer:
Explanation:
electrical field in vertical direction = V / .02 , .02 is distance between plate and V is potential difference .
force on electrons = V x e / .02
acceleration of falling electrons
= V x e /( .02 x m )
= 50 x 1.8 x 10¹¹ V
a = 9 x 10¹² V .
Time to cover length of plate
t = .05 / 10⁷ = 5 x 10⁻⁹ s , velocity of electron = 10⁷ m /s .
vertical distance covered by electron s = .01 m with acceleration a in time t
s = 1/2 a t²
.01 = .5 x 9 V x 10¹² x ( 5 x 10⁻⁹ )²
10⁻² = 4.5 x 25 x 10⁻¹⁸ x 10¹² V
V = 10⁴ / (4.5 x 25)
= 88.89 Volt .
What is it called when we have two gears connected by a chain in when large gear turns wants to small gear turns two or more times
A skier rides horizontally off of a 200 meter high cliff. If he lands 25 meters away from the base of the cliff, how fast was he skiing as he went off the edge?
A. 0.91 m/s
B. 2.91 m/s
C. 3.91 m/s
D. 25.91 m/s
Answer:
the answers, the correct one is C, v₀ₓ = vₓ = 3.91 m / s
Explanation:
This is a projectile launching exercise, in this case they indicate that when leaving the cliff it goes horizontally, therefore the initial vertical speed is zero, let's find the time to reach the base
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
at the base the height is zero (y = 0 m)
0 = y₀ + 0 - ½ g t²
t = √ (2y₀ / g)
we calculate
t = √ (2 200 / 9.8)
t = 6.389 s
with this time we calculate the horizontal speed
v₀ₓ = x / t
v₀ₓx = 25 / 6,389
v₀ₓ = vₓ = 3.91 m / s
When checking the answers, the correct one is C
Witch statement about electric force is true?
Question 13 (1 point)
The genetic material of an offspring of sexually reproducing organisms is best described as-
a
Ob
identical to that of the other offspring
a copy of the genetic material of the father.
a copy of The genetic material of the mother
genes from both parents, in unique combinations
ос
Od
The genetic material of an offspring of sexually reproducing organisms is best described as genes from both parents in unique combinations.
SEXUAL REPRODUCTION:
Sexual reproduction is a type of reproduction that involves two organisms (a male and a female). The male organism produces a gamete called SPERM while the female organism produces a gamete called EGG. The gametes are produced via a process called MEIOSIS. Meiosis is a process of cell division whereby genetically different daughter cells are produced. The genetic variation is attributed to a process called crossing over, which is the exchange of genetic material between non-sister chromatids of homologous chromosomes. At the end of the fertilization process between male sperm and female egg, the genetic material of the offspring will contain genes from both parents in unique combinations.Learn more at: https://brainly.com/question/11622266?referrer=searchResults
PLEASE HELP!!!!!!!!!
NEED HELP ASAP!!!!
problem is shown in the attached image
Answer:
im sorry but what is the question please tell me in the comments s i can help you
Explanation:
During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m high. When the ball reaches the net,
(a) does the ball clear it and
(b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net,
(c) does the ball clear it and
(d) what now is the distance between the center of the ball and the top of the net?
Answer:
a
Yes it clears
b
[tex]b= 0.19 \ m[/tex]
c
No it does not clear
d
[tex]z= 0.86 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]
The height of the ball above the ground is [tex]h = 2.3 7 \ m[/tex]
The distance of the net is [tex]d = 12 \ m[/tex]
The height of the net is [tex]H = 0.9 \ m[/tex]
Generally the time taken for the ball to reach the net is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
=> [tex]t = \frac{12}{23.6}[/tex]
=> [tex]t = 0.508 \ s[/tex]
Generally the change in height of the ball after t is mathematically represented as
[tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]
Here u is the initial velocity which is zero given that the ball was at rest initially
So
[tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]
=> [tex]\Delta h =1.28 \ m[/tex]
Generally the new height of the ball is mathematically evaluated as
[tex]s= h-\Delta h[/tex]
=> [tex]s = 2.37 - 1.28[/tex]
=> [tex]s = 1.09 \ m[/tex]
From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]b = s - H[/tex]
=> [tex]b = 1.09 - 0.90[/tex]
=> [tex]b= 0.19 \ m[/tex]
Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal , the velocity along the x-axis is
[tex]v_x = v cos(5)[/tex]
=> [tex]v_x = 23.6 cos(5)[/tex]
=> [tex]v_x = 23.5 \ m/s[/tex]
The velocity along the y-axis is
[tex]v_y = v sin(5)[/tex]
=> [tex]v_y = 23.6 sin(5)[/tex]
=> [tex]v_y = 2.06 \ m/s[/tex]
Generally the time taken for the ball to reach the net is
[tex]t = \frac{d}{v_x}[/tex]
=> [tex]t = \frac{12}{23.5}[/tex]
=> [tex]t =0.508 \ s[/tex]
Generally the change in height of the ball after t seconds is
[tex]c = v_yt + \frac{1}{2}gt^2[/tex]
=> [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]
=> [tex]c = 2.33[/tex]
Generally the new height of the ball after time t seconds is
[tex]e = h - c[/tex]
=> [tex]e = 2.37 - 2.33[/tex]
=> [tex]e = 0.04 \ m[/tex]
From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]z = H-e[/tex]
=> [tex]z = 0.90 - 0.04[/tex]
=> [tex]z= 0.86 \ m[/tex]
(a) Yes, the ball clears the net.
(b) The distance between the center of the ball and the top of the net is 0.203 m.
(c) No, the ball does not clear the net.
(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.
What is a Projectile motion?When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.
Given,
The horizontal distance traveled by the ball is 12 m.
The height of the top of the net is 0.90 m.
The height of the horizontal launch of the ball is 2.37 m.
The time for the horizontal motion of the projectile that is the ball is,
[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]
The equation for the vertical motion of the projectile can be solved by substituting the above result.
[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]
Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.
The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]
The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]
For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,
[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]
Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,
[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]
The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.
[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]
Thus, the distance between the center of the ball and the top of the net is -2,32 m.
When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.
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The space shuttle slows down from 1200 km/hr to 400 km/hr over 10 seconds. What is the space shuttles deceleration?
That's if the answer should be in m/s2.
Can two spheres of different diameters and different
masses have the same moment of inertia?
Answer:
NO
Explanation:
I=mr^2 this means moment inertia depends upon mass and square of radius or distance.