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a_sh-v [17]
2 years ago
10

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

ach block, as shown above. The graph shows how F varies with time t. Which block has the greatest average power provided to it between t = 0 s and t = 3 s?
Physics
1 answer:
Anon25 [30]2 years ago
7 0

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>
  • This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

  • F is the applied force
  • v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

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The electrons of carbon and hydrogen are _____________ in this molecule of methane.
nlexa [21]

Answer:

The answer is C

7 0
3 years ago
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
3 years ago
Which of the following is an SI base unit for measuring length?
irina [24]
Meters it the SI unit for measuring length.
6 0
3 years ago
Read 2 more answers
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
What do astronomers use in addition to parallax to find the actual distance of stars that are close to Earth?
IgorLugansk [536]

Answer:

trigonometry (guessing)

Explanation:

ellipse: is the shape of an orbit : looks like an oval

periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth

parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth

https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

7 0
2 years ago
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