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Sati [7]
4 years ago
13

Which letter represents the direction of the centripetal force?

Physics
1 answer:
AlladinOne [14]4 years ago
6 0
I think there should be a picture。
But the direction of the centripetal force is towards the center。
Hope that helps。
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What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie
leonid [27]

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)  

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

8 0
3 years ago
Read 2 more answers
The low-frequency speaker of a stereo set has a surface area of 0.06 m2 and produces 2.03 W of acoustical power. What is the int
zimovet [89]

Answer:

33.83W/m²

Explanation:

The intensity of the speake at the surface is

I = P/A

I = 2.03W / 0.06m²

I = 33.83W/m²

8 0
3 years ago
A rocket ship starts from rest and turns on its forward booster rockets causing it to have a constant acceleration of 4 meters p
bagirrra123 [75]

Complete question is;

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?

Answer:

v = 12 m/s

Explanation:

We are given;

Initial velocity; u = 0 m/s (because ship starts from rest)

Acceleration; a = 4 m/s²

Time; t = 3 s

To find velocity after 3 s, we will use Newton's first equation of motion;

v = u + at

v = 0 + (4 × 3)

v = 12 m/s

6 0
3 years ago
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

7 0
4 years ago
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