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lyudmila [28]
3 years ago
15

Which one of Newton's Laws fits this statement:

Physics
1 answer:
BabaBlast [244]3 years ago
5 0

Answer:

First Law

Explanation:

I learned this in 6th grade

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Mickey had three substances. He hit each of them with a hammer. A red
Verizon [17]

Answer:

A & D

Explanation:

:p

6 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
3 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
A tablet is made by crushing several chemicals, combining the chemicals as powders, and pressing the powders together to make a
slavikrds [6]

Chemical change , I think
3 0
3 years ago
Read 2 more answers
Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a
xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, h_b.

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, h_d. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be h_d and the distance travelled by the ball measured from the top be h_b.

It follows that h_d+h_b=9.8.

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

h_b=0\times t + 0.5\times 9.8t^2 (since v_{b0} =0.)

h_b=4.9t^2

Dart:

h_d=17.8\times t - 0.5\times9.8t^2 (the acceleration is opposite to the displacement, hence the negative sign)

h_d=17.8\times t - 4.9t^2

But

h_b+h_d =9.8

17.8\times t - 4.9t^2+4.9t^2 =9.8

17.8\times t = 9.8

t = 0.55

The height of the collision is the height of the dart above the ground, h_d.

h_d=17.8\times t - 4.9t^2

h_d=17.8\times 0.55 - 4.9\times(0.55)^2

h_d=9.79 - 1.48225

h_d=8.3

8 0
3 years ago
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