We know
[tex]\boxed{\sf a=\dfrac{v-u}{t}}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{v-u}{a}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{18-80}{-16}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{-62}{16}[/tex]
[tex]\\ \sf\longmapsto t=3.8s[/tex]
A 1.5 kg ball has a velocity of 12 m/s just before it strikes the floor. Find the impulse on the ball if the ball bounces up with a velocity of 10 m/s.
Hi there!
Recall:
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Let the direction TOWARDS the floor be POSITIVE, and AWAY be NEGATIVE.
Plug in the givan values:
Δp = 1.5(-10 - 12) = -33 Ns
**OR, the magnitude: |-33| = 33 Ns
This property of waves is the only property where the relationship between energy and this property are indirect or inverse
Answer: I don't understand
Explanation:
study and pay attention
scientific notation. jst 11a
Answer:
2 * 10^4
Explanation:
(3 * 10^4)(4 * 10^4) / (6 * 10^4)
= (12 * 10^8)/(6 * 10^4)
= 2 * 10^4
A toaster oven indicates that it operates at 1500 W on a 110 V
circuit. What is the resistance of the oven?
[tex]P=U.I => I=\frac{P}{U}=\frac{1500}{110}=\frac{150}{11}<A>\\I=\frac{U}{R}=> R=\frac{U}{I} = \frac{110}{\frac{150}{11} }=8.06< ohm>[/tex]
The answer is: A. 8.06 ohm
ok done. Thank to me :>
Bill pushed 327 kg. bucket of concrete with a force of 10N. What was the
acceleration?
Answer:
F(10)=mass(327)x acceleration(m/s)
Explanation:
A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in energy of the system
Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much
Find the area of the given figure
Answer:
77 sq.cm
Explanation:
Solution,
Trapezium
Perpendicular (P1)=8cm
Perpendicular(P2)=14cm
height(h)=7cm
Now,
We know that,
Area(A)= 1/2 ×h (P1+P2)
= 1/2 × 7cm (8cm+14cm)
= 1/2 × 7cm(22cm)
= 1/2 × 154 squarecm
=77 sq.cm
An object, initially at rest, moves 475 m in 19 s. What is its acceleration? *
[tex]\\ \sf\Rrightarrow s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf\Rrightarrow s=0(19)+\dfrac{1}{2}a(19)^2[/tex]
[tex]\\ \sf\Rrightarrow 475=\dfrac{361a}{2}[/tex]
[tex]\\ \sf\Rrightarrow 950=361a[/tex]
[tex]\\ \sf\Rrightarrow a\approx 2.3m/s^2[/tex]
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
[tex]v=\sqrt{\frac{GM}{R}}[/tex]
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
[tex]v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}[/tex]
v = 7660.25 m/s
Learn more about orbital velocity here:
https://brainly.com/question/541239
A 300 g ball swings in a vertical circle at the end of a 1.3-m-long string. When the ball is at the bottom of the circle, the tension in the string is 13 N.
What is the speed of the ball at that point?
Answer:
0.23N for the speed
at the bottom of the circle
Two 3.0g bullets are fired with speeds of 40.0 m/s
and 80.0 m/s, respectively. What are their kinetic
energies? Which bullet has more kinetic energy? What is the ratio of their kinetic energies?
[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]
Kinetic energy of the 1st bullet [tex] \sf = \frac{1}{2} \times m1 \times {(v1)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(40)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 1600J \\ \sf = 2.4J[/tex]Kinetic energy of the 2nd bullet [tex] \sf = \frac{1}{2} \times m2 \times {(v2)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(80)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 6400J \\ \sf = 9.6J[/tex] So, the 2nd bullet which has greater velocity has more kinetic energy.Therefore, the ratio of their kinetic energies[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]
Answer:
The kinetic energies of the bullets are 2.4 J and 9.6 J.
The bullet having greater velocity has more kinetic energy.
The ratio of their kinetic energies is 1 : 4.
Hope you could get an idea from here
Doubt clarification - use comment section.
The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.
The ratio of the kinetic energy of the bullets is 1:4.
What is kinetic energy?The kinetic energy of an object is the energy possessed by the object due to its motion.
The kinetic energy of the two bullets is calculated as follows;
[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]
Ratio of the kinetic energy of the bullets is calculated as follows;
[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]
Learn more about kinetic energy here: https://brainly.com/question/25959744
The practice of science can answer only scientific questions. And scientific questions guide the design of investigations. What must be true of the possible answers to a scientific question? A. They are popular with a majority of scientists. B. They agree with all prior experiments. O C. They can be supported by evidence. O D. They lead to increased funding of scientific research.
Answer:
They must be supported by evidence.
Explanation:
Every scientific theory or scientifc claim must have scientific evidence.
A 10.0-kg crate slides along a raised horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second, roughened horizontal surface as shown in the figure. What is the kinetic energy of the crate as it reaches the lower surface?
The kinetic energy of the crate as it reaches the lower surface is 80 J
Kinetic energy is the energy possed by an object in motion. Mathematically, the kinetic energy can be expressed as follow:
KE = ½mv²
With the above formula, the kinetic energy of the crate can be obtained as follow:
Mass (m) = 10 KgVelocity (v) = 4 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 10 × 4²
KE = 5 × 16
KE = 80 J
Therefore, the kinetic energy of the crate is 80 J
Learn more about kinetic energy: https://brainly.com/question/14449170
A student on her way to school walks four blocker east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than, greater than, or the same?
Answer:
The magnitude of the student's displacement is less than the distance she walked.
Walking: 11 Blocks
Displacement: 8.54 Blocks
Explanation:
See the attached diagram. The unit of length is blocks. We can add the actual blocks walked as shown. She walked a total of 11 blocks.
Her displacement is the distance measured directly from where she started (line A). Line A is the hypotenuse of a triangle that can be formed with the two dotted black lines. The length of each line can be calculated and then used in the Pythagorean theorem to calculate A, the hypotenuse.
That result is 8.54 Blocks, a shorter distance, once she earns her wings.
If a student on her way to school walks four blockers east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than the total distance walked by her.
What is displacement?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard.
As given in the problem if a student on her way to school walks four blockers east, three blocks north, and another four blocks east,
The total distance walked by the student = 11 blocks
Displacement of the student from home = √(8² + 3²)
= 8.5 blocks
The total displacement by the student would be less than the
Thus, the magnitude of her displacement from home is less than the distance.
To learn more about displacement here, refer to the link given below ;
https://brainly.com/question/10919017
#SPJ2
define the unit of work
[tex]\text{The unit of work is same as energy. Newton-meters(Nm) or Joules(J).}\\[/tex]
what is the affect of density of air in the velocity of sound??
Answer: Hello!
The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to m, the mass of the oscillating object. The speed of sound in air is low, because air is easily compressible.
Explanation:
Mark me brainest please. Hope I helped! Hope you make an 100% Anna♥
Please sub to Addie Nahoe! Bye!♥
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32. 4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate? The wave has traveled 32. 4 cm in 3 seconds. The wave has traveled 32. 4 cm in 9 seconds. The wave has traveled 97. 2 cm in 3 seconds. The wave has traveled 97. 2 cm in 1 second.
The true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
In Physics, we define a wave as a disturbance along a medium that transfers energy. The wavelength of a wave is the distance covered by the wave while the frequency of the wave is the number of cycles of the wave completed per second.
The period of the wave is the inverse of the frequency of the wave. It is defined as the time taken for the wave to complete a cycle and it is measured in seconds.
The wave formula is given as;
v = λf
v = velocity of the wave (distance traveled by the wave in one second)
λ = wavelength of the wave
f = frequency of the wave
So;
λ = 32.4 cm
f = 3 hertz
v = 32.4 cm × 3 hertz
v = 97. 2 cms-1
Hence, the true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
Learn more: https://brainly.com/question/14588679
7. If the speed of light in medium-1 and medium-2 are 2.5x 10 m/s and 2x 10 m/s respectively then the refractive index of medium-1 with respect to medium-2 is (a) 3/2.5 (b) 2/2.5 (c) 2.5/3
Answer:
Answer to the question is 4 / 5
To know the answer with proof see the above attachmentThe basic SI unit of length is
Answer:
m
Explanation:
Metres, m, is the SI unit of length
Hope this helped you-have a good day bro cya)
Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.
(a) How far did he travel?
(b) What was his average speed?
(c) What was his displacement?
(d) What was his average velocity?
Answer:
Explanation:
See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.
The individual driving times and distances are summed to provide:
(a) How far did he travel?
103 km
(b) What was his average speed?
12.88 km/h
(c) What was his displacement?
73.82 km
(d) What was his average velocity?
9.228 km/h
An eagle flying at a constant 120 km/h and has kinetic energy of 2,800 J. What is the mass of the eagle?
Answer:
The mass of the eagle is about 5 Kg
Explanation:
1/2 M= Ke/V^2
120 km/h = 33.3333m/s
1/2 M = 2,800/33.3333^2
1/2 M = 2,800/ 1111.10888889
1/2 M = 2.52000504001
(2) 1/2 M = (2) 2.52000504h001
= M = 5.04001008002
About 5 Kilograms
You plug your microwave into an outlet and then you heat up a piece of pizza in it. This is an example of an...
Microwave Radiation????????????
A 3-column table with 1 row. The first column titled distance travelled (meters) has entry 6. 1. The second column labeled lower track elapsed time (seconds) has entry 4. 92. The third column labeled higher track elapsed time (seconds) has entry 3. 36. Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the higher track? How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is. Calculate speeds for each track. How much faster was the car on the higher track than the lower track?.
Answer:
B,A,A
Explanation:
Answer:
Other guy is correct b,a,a
Explanation:
The diagram shows a nephron.
A nephron. W points to branch of renal artery. X points to branch of renal vein. Y points to tubular point of nephron. Z points to collecting duct.
Where is the blood first filtered?
Answer:
w
Explanation:
renal artery...the blood flow into the kidney via this blood vassles
and the filtration take place is called ultrafiltration
i. a A ball is released from a height of 45 m top of the building. If it strikes the ground surface after 5 seconds, calculate the acceleration and final velocity of the ball. [18 m/s, 3.6 m/s²].
Answer:
3.6m/s², 18m/s
Explanation:
So according to h=1/2at², 45=1/2a*25, a=3.6m/s² (which means that it didn't happen on Earth?)
Final velocity = at = 3.6*5 = 18 m/s
1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)
Hypothesis:?
Conclusion: ?
As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
= Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is
= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons
who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation
Answer:
electricity because it has more percentage nd energy
Explanation:
mark me brainliest pl
What is the work done in lifting 60 kg of blocks to a height of 20m
Answer:
The answer is 1200
Explanation:
A car which is traveling at a velocity of 27 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 430 m. How fast is it going after that acceleration?
73.9m/s (1dp)
1) list everything that you are given using suvat, where s is distance, u is initial velocity(speed), v is final velocity(speed), a is acceleration and t is time
s = 430m
u = 27 m/s
v = ? (we need to work out)
a = 5.5m/s^2
t = (we are not given this value)
2) use an equation that doesn't involve the time
[tex] {v}^{2} = u {}^{2} + 2as[/tex]
3) input the values that we have
[tex] {v}^{2} = ( {27})^{2} + 2(5.5)(430)[/tex]
[tex]v {}^{2} = 5459[/tex]
[tex]v = \sqrt{5459} = 73.9[/tex]
so the answer is 73.9m/s to 1dp
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
What is the efficiency of the lamp?
С
18%
D
85%
A
0.15%
15%
B
Answer:
[tex]15\%[/tex].
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is [tex]40\; \rm W[/tex] (same as [tex]40\; \rm J \cdot s^{-1}[/tex],) meaning that [tex]40\; \rm J[/tex] of energy is supplied to this lamp every second.
The question states that [tex]34\; \rm J[/tex] out of that [tex]40\; \rm J[/tex] of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the [tex](40\; \rm J - 34\; \rm J) = 6\; \rm J[/tex] of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive [tex]40\; \rm J[/tex] of energy input and would outputs [tex]6\; \rm J[/tex] of useful work. The efficiency of this lamp would be:
[tex]\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}[/tex].
