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3 years ago

5

A bug flies at a velocity of 0.75 m/s into an oncoming breeze blowing at 0.25 m/s. What is the resultant velocity of the bug?

1 answer:

goldenfox [79]3 years ago

3 0

C: .50 m/s is the answer

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If a proton were released in the electric field above, what direction would it move?

Afina-wow [57]

Answer:

d because the proton would move towards the negative plate

Explanation:

5 0

3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

3 years ago

The cable of a crane is lifting a 950 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/sin a distance of 5.5m

Umnica [9.8K]

Explanation:

a) How much work is done by gravity?

w = f x d
w = 950 x 10 x 5.5 = 52250j

b) How much work is done by tension?

v²=u²+2as
0.75²=0.25²+2a x5.5
0.56=0.06+2a x5.5
2a x5.5 = 0.56 - 0.06
2a x 5.5 =0.5
11a=0.5
a = 0.5/11 = 0.05m/s²

w = f x d

w = 950 x 0.05 x 5.5 = 261.25j

7 0

3 years ago

A single conducting loop of wire has an area of 7.26E-2 m2 and a resistance of 117 Ω. Perpendicular to the plane of the loop is

cricket20 [7]

Answer:

$\frac{dB}{dt}$ = 591.45 T/s

Explanation:

i = induced current in the loop = 0.367 A

R = Resistance of the loop = 117 Ω

E = Induced voltage

Induced voltage is given as

E = i R

E = (0.367) (117)

E = 42.939 volts

$\frac{dB}{dt}$ = rate of change of magnetic field

A = area of loop = 7.26 x 10⁻² m²

Induced emf is given as

$E = A\frac{dB}{dt}$

$42.939 = (7.26\times 10^{-2})\frac{dB}{dt}$

$\frac{dB}{dt}$ = 591.45 T/s

3 0

3 years ago

the two ropes are used to vertically lower a 255 kg piano from exactly 4 m form a seocnd sotry window to the ground how much wor

Ilya [14]

Complete Question

The Question diagram is attached below

Answer:

a) $W_{Fg}= 12500 Nm$

b) $W_{T_1}= - 6339.3Nm$

c) $W_{T_2}= - 3662.8Nm$

Explanation:

From the question we are told that:

Mass $m=255kg$

Distance $d=4m$

Generally the equation for Work done is mathematically given by

$W=F*d$

For $F_g$

$W_{Fg}=2500 x 5.3$

$W_{Fg}= 12500 Nm$

For $T_1$

$W_{T_1}= - {1830 sin(60) x 4}$

$W_{T_1}= - 6339.3Nm$

For $T_2$

$W_{T_2}= - {1295 sin(45) x 4}$

$W_{T_2}= - 3662.8Nm$

6 0

3 years ago