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rodikova [14]
3 years ago
12

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
arsen [322]3 years ago
5 0

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates  to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m=1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}=7*10^{14} m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

X=u[2y/a]^{1/2}

=4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}

=3*10^{-2}= 3 cm

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49. A block is pushed across a horizontal surface with a
nata0808 [166]

Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

kinetic friction coefficient = 0.15

(a) acceleration, a = 2.53 m/s^2

According to the newton's second law

Net force = mass x acceleration

F - friction force = m a

150 - 0.15 x m g = m a

150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4

8 0
3 years ago
A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s
AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight W = 150 N

Length of rope r = 4.1 m

Initial speed of ball v = 5.5 \frac{m}{s}

For finding the tension in the rope,

First find the mass of rod,

mg = 150                          ( g = 9.8 \frac{m}{s^{2} } )

  m = \frac{150}{9.8}

  m = 15.3 kg

Tension in the rope is,

  T = mg + \frac{mv^{2} }{r}

  T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}

  T = 262.88 N

Therefore, the tension in the rope is 262.88 N

7 0
3 years ago
A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2, At equilibrium, the f
Alexxandr [17]

Answer:

13 530 482

Explanation:

                            H2    +          I2     ------>      2HI

start (mol)             0.3785         0.3818                   0

change (mol)       -0.3534        -0.3534            +0.7067

equilibrium (mol)  0.0251         0.0284             0.7067

concentra (mol/L) 0.0068        0.0077              0.1926

K_{c} = \frac{0.1926^{2}}{0.0068^{2}*0.0077^{2} } = 13530482

7 0
3 years ago
Does gravity hold things together?
mariarad [96]
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3 0
3 years ago
Read 2 more answers
A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to
zysi [14]

Answer:

50.2 m

Explanation:

We can solve the problem by using the following SUVAT equation for the vertical position of the rock:

y(t)=h+ut+\frac{1}{2}gt^2

where

h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position

u = 0 is the initial velocity of the rock

t is the time

g=-9.8 m/s^2 is the acceleration of gravity

When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon:

h=\frac{1}{2}gt^2 = -\frac{1}{2}(-9.8)(3.2)^2=50.2 m

5 0
3 years ago
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