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lapo4ka [179]
3 years ago
6

Find the force on an object which has a mass of 20 kg and an acceleration of 10 m/s2.

Physics
1 answer:
Ilya [14]3 years ago
3 0

Answer:

200N

Explanation:

mass = 20 kg

acceleration = 10 m/s^2

Force = mass * acceleration

         = 20 *10

         = 200 N

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Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a
bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>
  • This is a single force that balances other given forces.

The given parameters:

  • First force, F₁ = 10 N
  • Second force, F₂ = 10 N
  • Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0
2 years ago
Identify the palindromic base sequence in the DNA donor molecule shown.
Airida [17]

Explanation:

Since a double helix is formed by two paired antiparallel strands of nucleotides that run in opposite directions, and the nucleotides always pair in the same way (adenine (A) with thymine (T) in DNA or uracil (U) in RNA; cytosine (C) with guanine (G)), a (single-stranded) nucleotide sequence is said to be a palindrome

6 0
2 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
3 years ago
Which of the below is an example of mimicry that enables prey species avoid predation?
Marta_Voda [28]

b..a harmless organism imitating the look of a harmful organism

Explanation:

A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

Prey are smaller and less harmful organisms often hunted by larger organisms usually carnivores.

Mimicry is a form of evolutionary adaptation process in which two organisms of the same specie or different species tends to look alike.

It is a subtle defense mechanism developed by organism over an extended period of time.

learn more:

Adaptation brainly.com/question/11105547

#learnwithBrainly

6 0
3 years ago
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