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nasty-shy [4]
2 years ago
15

Lahat ay tumutukoy sa bugtong, maliban sa:

Physics
1 answer:
AlladinOne [14]2 years ago
5 0

Answer:

hsksjshjajskslksbZuusbs

js

Explanation:

jaksvshsuzbsbsbsbzmzlKsbsbjzbzzbhzjsbsjsisksuzbbzznkskskwwhhsyshsbsjskslsksuwhsbbskslzkzjzhzbzbsbssnhahahahshhahahahshahshahsjshhshahahahahhasorrrrrryyyyyy po hindi kopo alam ang isasagot ✌️sksvsjsish

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What is the abbreviation for mojave desert?
vagabundo [1.1K]
There is no abbreviation
3 0
3 years ago
An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?​
valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2)    (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

5 0
2 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an e
PIT_PIT [208]

Answer:

E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

Explanation:

From the question we are told that

   The charge on the small object is Q = -4.00 \ nC =  -4.00 *10^{-9} \  C

   The force is  F  =  24  \ nN  =  24 *10^{-9} \  N

    Generally the magnitude of the electric  field is mathematically represented as

       E  =  \frac{F}{Q}

=>    E  =  \frac{ 24 *10^{-9}} {-4 *10^{-9 }}

=>     E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

 

6 0
2 years ago
A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling
Tasya [4]

Answer:

Solving for time :

(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known

2. The unknown constant that we want to solve)

s = (1/2)(u+v)t <--- one of the formulas

from linear motion

s (distance) = 0.05m

u (initial velocity) = 100m/s

v (final velocity) = 0 m/s (it stops)

t (time taken for change in velocity) = to be found

0.05 = (1/2)(100+0)t

t = 0.001 seconds

Solving for the resistant force :

Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.

When the bullet stops :

F net = 0

F r = F imp

F r = (mu -mv)/t

F r = (0.01x100-0.01x0)/0.001

F r = 1/0.001

F r = 1000N

4 0
3 years ago
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