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ra1l [238]
3 years ago
12

Terms used to measure motion

Physics
2 answers:
pychu [463]3 years ago
7 0

Answer:

velocity term

Explanation:

krek1111 [17]3 years ago
7 0

Answer:velocity term

Explanation:

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Why do mirrors form inverted images?
marysya [2.9K]

Answer:

Explanation:

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted).

3 0
3 years ago
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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
What happens when the object is placed at F? Explain your answer.
sp2606 [1]

Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.

6 0
3 years ago
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A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the
belka [17]
We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values, 
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶)  / (3 * 10⁻²)²

F = 100 N  [ East of positive charge ]

Hope this helps!
5 0
4 years ago
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If a force of 0.05 N acts on a body of mass 100 grams. Determine the acceleration at which that body moves.​
eduard
Newton’s second law is a=F/m this is what we will be using to solve this
However first you need to convert g to kg
100g= 0.1kg

0.05/0.1=0.5 m/s^2
3 0
3 years ago
Read 2 more answers
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