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2 years ago

A swimmer swims 110 m toward the shore in 72 s. What is the swimmers velocity in neters per second?

Physics

1 answer:

MrRissso [65]2 years ago

6 0

Answer:

1.53 m/s

Explanation:

Velocity is given in meters per second, so if it takes the swimmer 72 second to swim 110 meters:

m/s = 110 meters / 72 seconds = 1.527 or 1.53 m/s

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Which of these accurately describes the products of a reaction?

stiv31 [10]

The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O

hope this helps

3 0

2 years ago

Read 2 more answers

A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the

iren [92.7K]

Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

Power(P)= Work(W)/Time taken(T)

4 0

3 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0

3 years ago

A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds

Inessa05 [86]

Answer:

A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel

20miles / 5sec = 4miles /sec would be the average speed for the last 20 m

Explanation:

The answer is 4 m/s.

In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).

The relation of speed (v), distance (d), and time (t) can be expressed as:

v = d/t

We need to calculate the speed of the second 5 seconds of the travel:

d = 20 m (total 30 meters - first 10 meters)

t = 5 s (time from t = 5 seconds to t = 10 seconds)

Thus:

v = 20m / 5s = 4 m/s

PLEASE GIVE BRAINIEST!! HOPE THIS HELPS

5 0

3 years ago