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azamat
2 years ago
14

HELP ASAAPP!!!! if A MAN JAY WALKED ACROSS THE STREET IS THAT BAD

Physics
1 answer:
kaheart [24]2 years ago
4 0
True, yes it is bad to jay walk across the street
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Could someone please help me
BaLLatris [955]

Answer:

the size of the shadow will be smaller due to smaller hands

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3 years ago
A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in-
Leto [7]

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

4 0
3 years ago
Your image appears 3.0 m from a plane mirror. How far IS YOUR IMAGE IN<br> RELATION TO YOU?*
Vsevolod [243]

Answer:

Being a plane mirror the Image is formed 3 metres beyond the mirror . So total distance is 3+3 = 6metres

8 0
3 years ago
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How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?
kozerog [31]
F = ma,    where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.
5 0
3 years ago
Read 2 more answers
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
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