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Ivanshal [37]
2 years ago
10

PLS ANSWER FAST WILL GIVE BRAINLY!!!! ANSWER ONLY IF YOU KNOW!!!!

Physics
1 answer:
timurjin [86]2 years ago
5 0

Answer:

The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons. These opposite charges cancel each other out making the atom neutral.

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Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
Describe all the characteristics of Earth that make it ideal for living creatures.
Lady_Fox [76]
Earths atmosphere, Temperature, Plant life and Oxygen I guess
8 0
3 years ago
Read 2 more answers
At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit
matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

       

Starting point. With the mouse in the center

            L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

         I = 1/3 M L²

Final point. When the mouse is at the end of the rod

          L_{f} = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

        L₀ = L_{f}

        I w₀ = (I + m L²) w

        w = I / I + m L²) w₀

We substitute the moment of inertia

        w  = 1/3 M L² / (1/3 M + m) L²    w₀

        w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

      w = 1/3 / (1/3 + 0.02) w₀

      w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

        w = 0.943 rad / s

3 0
3 years ago
Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction
Pavel [41]

Answer:

3.2\cdot 10^{-8} N

Explanation:

The inital electrostatic force between the two spheres is given by:

F=k\frac{q_1 q_2}{r^2}

where

F=8\cdot 10^{-9} N is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

r'=\frac{r}{2}

Therefore, the new electrostatic force will be

F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F

So the force has increased by a factor 4. By using F=8\cdot 10^{-9} N, we find

F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N

6 0
3 years ago
A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a
Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

F=ma      Eqn. (1)

where

F : is the Net Force in Newtons ( N )

m : is the mass ( kg )

a  : is the acceleration ( m/s^2 )

We also know that the acceleration is denoted by the velocity ( v ) of an object as a function of time ( t ) with

a=\frac{v}{t}         Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

F=m\frac{v}{t}\\ \\Ft=mv     Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. Ft is in fact the Impulse  J of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ 22N and t=4 sec (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

J=Ft\\J=(22N)(4sec)\\J=88 kg m/s

So the impulse of the cart is J=88kg m/s

Then, we know that the cart is moving at v=10 m/s. Plugging in the values in Eqn. (3) we have:

(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg

So the mass of the cart is m=8.8kg.

8 0
3 years ago
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