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3 years ago

Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?

Physics

1 answer:

Jet001 [13]3 years ago

7 0

I think D x=vxt because it's equation finding change of x (displacement) and using time

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Can I improve the design of my simple machine? How?

blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

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3 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

Read 2 more answers

Which statement is true about an atom and an element?

Lemur [1.5K]

"<span>An atom is the smallest unit of matter and an element is a pure substance that is made of identical atoms" is correct. Although atoms can be broken down further now, it still take a whole atom to make an element. </span>

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3 years ago

In transistor emitter current is equal to which current?

Paladinen [302]

In transistor,
Emitter current is equal to the sum of base current and collector current.
Thanks!

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3 years ago

A motorbike reaches a speed of 20 m/s over 60m, whilst

Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

An equation of uniformly accelerated motion

V = vo + at

Vt² = vo² + 2a (x-xo)

x = distance on t

vo / vi = initial speed

vt / vf = speed on t / final speed

a = acceleration

vf=20 m/s

d = 60 m

a = 3 m/s²

$\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s$

7 0

4 years ago