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3 years ago

Examine the scenario.

2 answers:

8 0

The answer is "A magnetic field influences the motion of charged particles. The negatively-charged particles will experience a force acting against their motion, so the negatively-charged particles would slow down and stop moving."

This is because for example;

Have you ever seen a magnet? If so, you would experience the same scienario as above.

aniked [119]3 years ago

5 0

Answer:

option (c)

Explanation:

According to the Fleming's left hand rule, if we spread out thumb, fore finger and middle finger of our left hand mutually perpendicular to each other, then thumb represents the direction of force, fore finger represents the direction of magnetic field and the middle finger represents the direction of current.

According to the question, positive charge particle moves in left wards direction, i.e., current is flowing towards left, the force is acting upwards, so the direction of magnetic field is in perpendicular outwards to the plane of paper.

When a negatively charged particle moves towards left , then the direction of force is downwards.

Thus, the magnetic field influences the motion of charged particles. The negatively-charged particles will experience a force downward because positive and negative particles would experience forces in opposite directions.

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Wheel A has a radius of R and wheel B has a radius of 2R. Wheel A has all the mass at the rim, while wheel B has the mass unifor

azamat

Answer:

Wheel A.

Explanation:

The lesser the moment of inertia, the greater the angular acceleration. Then, the moments of inertia of each wheel are described below:

Wheel A

$I_{A} = M\cdot R^{2}$

Wheel B

$I_{B} = \frac{1}{2}\cdot M \cdot (2\cdot R)^{2}$

$I_{B} = 2\cdot M\cdot R^{2}$

The wheel A accelerates faster in response to the torque.

3 0

3 years ago

Definition of waves

Dmitriy789 [7]

Answer:

A wave can be described as a disturbance that travels through a medium from one location to another location.

Explanation:

6 0

3 years ago

Distinguishing between Speed and Velocity.

koban [17]

Answer:

Speed: Distance per time, 400 km/h, and a scalar quantity.

Velocity: Displacement per time, 20 m/s south, and a vector quantity.

Explanation:

Hope this helps! Please mark as brainliest.

Thanks!

8 0

3 years ago

A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t :

Equations of the uniformly accelerated rectilinear motion of upward (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the initial vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 = (v₀y)²

$v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }$

v₀y = 14.3 m/s

Calculation of the maximum height the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball (R)

We replace data in the equation (1)

x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

$v= \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v= \sqrt{(7.7)^{2}+ (-14.3)^{2} }$

v= 16,2 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })$

$\alpha = tan^{-1} (\frac{-14.3 }{7.7 })$

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0

3 years ago

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

$f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz$

3 0

3 years ago