1.
In 10 seconds a car accelerates 4m/s? to 50 m/s.
How fast was the car going before it accelerated?
2.
Variables:
Equation and Solve:
In 10 seconds a car accelerates 4m/s2 to 50 m/s. How fast was the car going before it accelerated?
Answer:
i love devon bostick
Explanation:
cause....... i do
Answer: It was going 4 m/s
Explanation:
how many current can absorb human body
Answer:
about 10 milliamperes
Explanation:
currents above 10ma can paralize or "freeze bodies"
During a basketball game a kid tries to show off his fancy dribbling skills by going 3.5 m east and 5 m north. Dan is way too good to follow him through this unnecessary display. What magnitude and direction could Dan go to keep up with the other kid and only move in one straight line. Must show work for full credit
HELPP PLZZ
Answer:I litterly need a picture of this put the pic and Ill edit this question
Explanation:
What is the point on the globe to the right that receives 24 hours of daylight on the December solstice?
Answer: Explanation:
Locations above the Arctic Circle (north of 66.5 degrees latitude; 90 degrees minus the tilt of Earth's axis) receive 24 hours of sunlight.
Orbits: Sun
Need help on these 2 question ??hurry anyone please
Answer:
I know you may not be reading this but for number 13 the answer is 2 hours because t=d/s which is 4000/2000 which equals to 2.
Explanation:I've currently doing that problem as of the time i'm writing this.
define resistivity ?
what is the range of resistivity for metal and alloys
Answer:
The resistivity of an exceedingly good electrical conductor, such as hard-drawn copper, at 20° C (68° F) is 1.77 × 10-8 ohm-metre, or 1.77 × 10-6 ohm-centimetre. At the other extreme, electrical insulators have resistivities in the range 1012 to 1020 ohm-metres.
what are 5 different movements you can use to warm up
Answer:
1) Jumping Jacks (their a great way to get blood into your muscles and keep your blood running)
2) Toe touches (helps warm up your hamstring so you won‘t get injured easily)
3) Jog in place (Warm up leg muscles and whole body)
4) Arm circles (Warms up arm muscles)
5) 10,10,9,9. (This is something I like, you hop on 1 foot for 10 times and then switch to your other foot ten times without stopping, but after the ten you go back foryour previous hopping leg and hop 9 times this time and same for your other foot, THEN hop 8 times with the foot you started with and, keep going in that order in decending order until you reach 1, (a great way to warm up leg muscle)
Can someone please help me?
Answer:
1.0 A
Explanation:
Suppose that the net external force (push minus friction) exerted on a lawnmower is 51 N. parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?
Answer:
2.125 m/s^2
Explanation:
Use Newton's second law: Force equals mass times acceleration, F = ma. You know the force (51 N) and you know the mass of the mower (24 kg). Remembering that a Newton is the same as kg-m/s2, solve for acceleration:
F = ma
51 kg-m/s2 = 24 kg * a
2.125 m/s2 =a
If a net external force of 51N is extorted on a 24kg lawnmower parallel to the ground, the acceleration of the mower will be 2.125 m/s².
Given the data in the question;
Net External Force; [tex]F = 51N[/tex]
Mass of the Mower; [tex]m = 24kg[/tex]
Acceleration; [tex]a = ?[/tex]
To determine the acceleration of the mower, we make use of the equation from the Newton's Second Law of Motion:
[tex]F = m\ * \ a[/tex]
Where F is the force exerted, m is the mass and a is the acceleration.
But then, we know that; ( A newton is the same thing as [tex]1 kg.m/s^2[/tex] )
Hence, Our Net External Force; [tex]F = 51kg.m/s^2[/tex]
We substitute our values into the equation to find acceleration "a"
[tex]51kg.m/s^2 = 24kg\ *\ a\\\\a = \frac{51kg.m/s^2}{24kg} \\\\a = 2.125 m/s^2[/tex]
Therefore, If a net external force of 51N is extorted on a 24kg lawnmower parallel to the ground, the acceleration of the mower will be 2.125 m/s².
Learn more; https://brainly.com/question/12887896
Surfaces on the same body are assumed to experience meteorite impacts with the same frequency. If this is true, what could account for the stark contrast in impact crater density on bordering regions of Ganymede in the photo?
Answer:
The surface with fewer impact craters is a newer surface. It hasn’t been impacted as many times as the older surfaces
Explanation:
Sample Answer
Answer:
Newer surfaces are less damaged by impact craters, because they are less damaged by impact.
Explanation:
This is my answer.
what are the types of energy sources based on the time of replacement ?
Answer:
In general there are three sub-segments of "alternative" energy investment: solar energy, wind energy and hybrid electric vehicles
Which type of galaxy has arms that contain sites of active star formation and start close to a bulge in the center?
A. elliptical
B. irregular
C. barred spiral
D. normal spiral
Answer:D:normal spiral
Explanation:
Answer:
D
Explanation:
how is meiosis like binary fission?
How many types of joints are there in the human body? A. 4 B. 2 C. 12 D. 6
Answer:
there are six types of joints in the human body
Answer: The answer is D. 6
Explanation: Learned in health class a while ago.
If you know the answer please answer the following question down in the picture below.
Answer:
the answer is C
I hope I helped (◍•ᴗ•◍)❤
A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/sec2 . It runs out of fuel after 4 seconds and continue to rise, reaching a maximum height before falling back to Earth. a) Find the rocket's velocity and height at the moment fuel ends. b) Find the maximum this rocket can reach. c) Find the velocity the instant before the rocket crashes on the ground.
Answer:
(a). The rocket's velocity is 117.6 m/s.
(b). The rocket can reach at maximum height is 940.8 m.
(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.
Explanation:
Given that,
Acceleration = 29.4 m/s²
Time = 4 sec
(a). We need to calculate the rocket's velocity
Using equation of motion
[tex]v=u+at[/tex]
Put the value into the formula
[tex]v=0+29.4\times4[/tex]
[tex]v=117.6\ m/s[/tex]
We need to calculate the maximum height at the moment fuel ends
For the value of x₁
Using equation of motion
[tex]x_{1}=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]x_{1}=0+\dfrac{1}{2}\times29.4\times4^2[/tex]
[tex]x_{1}=235.2\ m[/tex]
We need to calculate the value of x₂
Using equation of motion
[tex]v^2=u^2-2gx_{2}[/tex]
Put the value into the formula
[tex]0=u^2-2gx_{2}[/tex]
[tex]x_{2}=\dfrac{u^2}{2g}[/tex]
Put the value in to the formula
[tex]x_{2}=\dfrac{117.6^2}{2\times9.8}[/tex]
[tex]x_{2}=705.6\ m[/tex]
(b). We need to calculate the maximum this rocket can reach
Using formula for height
[tex]H=x_{1}+x_{2}[/tex]
Put the value into the formula
[tex]H=235.2+705.6[/tex]
[tex]H=940.8\ m[/tex]
(c). We need to calculate the velocity the instant before the rocket crashes on the ground
Using equation of motion
[tex]v^2=u^2+2gh[/tex]
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times940.8}[/tex]
[tex]v=135.7\ m/s[/tex]
Hence, (a). The rocket's velocity is 117.6 m/s.
(b). The rocket can reach at maximum height is 940.8 m.
(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.
Which graph shows the change in velocity of an object in free fall?
Answer:
C!
Explanation:
I just did the test !
Please stay safe and wear a mask , we are in the mist of a pandemic ! !
The graph shows the change in velocity of an object in free fall is according to option C.
What is free falling?When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.
When an object is made to fall freely, it's velocity is only depended on the acceleration due to gravity, g. The velocity decreases till it become zero.
Thus, the correct option is C.
Learn more about free falling.
https://brainly.com/question/13299152
#SPJ5
What are the three components of the FITT acronym associated with overload and progression?
A.
frequency, intensity, and type
B.
time, frequency, and intensity
C.
fitness, intensity, and time
D.
type, intensity, and fitness
Answer:
B
Explanation:
What is the relationship between the structure of a white blood cell and its function in the body.
A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is
40m long, what is the ball's acceleration? Include only the number when
you put your answer.
Answer:
The acceleration of the snowball is 0.3125
Explanation:
The initial speed of the snowball up the hill, u = 0
The speed the snowball reaches, v = 5 m/s
The length of the hill, s = 40 m
The equation of motion of the snowball given the above parameters is therefore;
v² = u² + 2·a·s
Where;
a = The acceleration of the snowball
Plugging in the values, we have;
5² = 0² + 2 × a × 40
∴ 2 × 40 × a = 5² = 25
80 × a = 25
a = 25/80 = 5/16
a = The acceleration of the snowball = 5/16 m/s².
The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .
Calculate potential energy of a 10 kg object sitting on a 20 meter hill
Answer:
p.e=mgh
where m=10kg
g=10m/s(constant)
h=20m
p.e=10x20x10
p.e=2000joules
If a chemical reaction begins with 23 atoms of sodium, how much sodium will be present at the end of the reaction?
What is most occurring type of element on the periodic table?
Answer:
Hydrogen is the most abundant element in the Universe; helium is second. However, after this, the rank of abundance does not continue to correspond to the atomic number; oxygen has abundance rank 3, but atomic number 8. All others are substantially less common.
Explanation:
In the winter, people often buy large bags of rock salt to sprinkle on their walkways. Why do people do this?
Answer:
It helps melt the ice away on their walkway.
Explanation:
The salt makes the ice to freeze at a lower temperature. So when a solute (salt) comes into contact with the ice the freezing point is lowered.
A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope , its speed is 4 m/s. But 3 seconds later at the bottom of the slope , its speed is 22 m/s. What is the average acceleration?
Please Show Steps or work on how to do it :(( This topic confuses me and I just get lost.
What would the density of a liquid that has a mass of 275 grams and a volume of 200 mL?
Answer:
Mathematically: D = m/v. If you know what liquid you have, you can look up its density in a table. Once you know that, all you have to do to find the mass of the liquid is to measure its volume. Once you know density and volume, calculate mass using this relationship: mass = density • volume
Explanation:
The specific heat capacity of water is 4,200 J/kg °C.
Calculate the thermal energy that must be
absorbed by 1.4 kg of water to increase its
temperature by 25 °C.
help TvT
Why do storm go to Galveston?
Hurricanes and tropical storms can also spawn tornadoes and microbursts, create storm surges ... If you are unable to evacuate, go to your wind-safe room.
The unit we use to measure speed is
Answer:
However, to mention a few, there are other units of speed such as meters per second, feet per second, light-years per millennium, and knots. Feet per second, but specially meters per second are usually used to measure speed of animals, humans, and free fall objects. Knots is used to measure the speed of ship and/or boats.
Explanation:
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of friction of 2.7 N opposing the motion. what is the acceleration of the hockey puck?
Choices:
0.0054 m/s^2
0.005 m/s^2
200 m/s^2
190 m/s^2
Answer:
[tex]a=190\ m/s^2[/tex]
Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2[/tex]
So, the acceleration of the hockey puck is [tex]190\ m/s^2[/tex].
The magnitude of acceleration of the hockey puck is 190 m/s² .
Given data:
The mass of hockey puck is, m = 0.17 kg.
the magnitude of external force on the puck is, F = 35.0 N.
The magnitude of force of friction is, f = 2.7 N.
We need to find the net force acting on the puck first. For that consider the applied force in the x-direction. Then net force is,
F' = F - f
Solving as,
F' = 35.0 - 2.7
F' = 32.3 N
Now, use the Newton's Second law to obtain the magnitude of acceleration (a) as,
F' = ma
Solving as,
32.3 = 0.17 (a)
a = 32.3/0.17
a = 190 m/s²
Thus, we can conclude that the magnitude of acceleration of the hockey puck is 190 m/s² .
Learn more about the Newton's law of motion here:
https://brainly.com/question/13447525