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2 years ago

A cat accelerates from rest to 10m/s when it sees a dog. This takes 2 seconds. What was the acceleration of the cat

Physics

2 answers:

Nookie1986 [14]2 years ago

5 0

The formula is a= chance in velocity/time
A=10-0/2
A=10/2
A=5 m/s^2 (meters per second squared)

nlexa [21]2 years ago

4 0

Answer :

Acceleration of the cat is $a=5\ m/s^2$

Explanation:

It is given that,

Initial speed of the cat, u =0 (at rest)

Final speed of the cat, v = 10

Time taken, t = 2 s

Let a is the acceleration of the car. We know that the rate of change of velocity of an object is called its acceleration. Mathematically, it is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10-0}{2}$

$a=5\ m/s^2$

So, the acceleration of the cat was $5\ m/s^2$ . Hence, this is the required solution.

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Peter often gets sore muscles in the back of his lower legs from jogging what flexibility exercise could help him

nordsb [41]

I would say deep lunges

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2 years ago

Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you

kap26 [50]

Answer:

Explanation:

An atom is constructed of three different particles known as electrons, protons and neutrons.

These particles have different mass and charges and are responsible for various characters than an atom posses.

An electron has a negative charge, a proton has positive charge and charge of neutron is neutral. Equal number of electrons and protons are present in an atom that make it electrically neutral but different conditions can occur if we remove these particles from an atom.

1 : Model of an ionized atom - an ionized atom is one which has some net charge on this. It can be either a positive charge or a negative charge.

If we need to sketch the model of an ionized atom then one should either keep the number of electrons less or proton.

2: Model of radioactive atom : A radioactive atom is one an unstable atom and has access of energy in its center. It can be caused by adding either neutrons or protons.

3 0

2 years ago

A celebrating student throws a water balloon horizontally from a dormitory window that is 50 m above the ground. It hits the gro

Contact [7]

Answer:

a) The horizontal velocity of the balloon just before it hits the ground is 6 m/s

b) The magnitude of the vertical velocity of the balloon just before it hits the ground is 98 m/s.

Explanation:

Hi there!

The velocity and position vectors of the water balloon are given by the following equations:

r =(x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v =(v0x, v0y + g · t)

where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive) .

v = velocity vector at time t.

a) Please, see the attached figure for a graphic description of the problem.

Considering the origin of the frame of reference as the point of launch, notice that the position vector when the balloon hits the ground is

r1 = (60, -50) m

Then:

r1x = 60 m = v0x · t

r1y = -50 m = 1/2 · (-9.8 m/s²) · t²

(notice that the initial vertical velocity is zero, see figure).

Solving r1y for t:

(-50 m · 2) / -9.8 m/s² = t²

t = 10 s

Now, let´s replace t in the r1x equation and solve it for the horizontal component of the velocity:

60 m = v0x · 10 s

v0x = 60 m / 10 s

v0x = 6 m/s

The initial horizontal component of the velocity is 6 m/s. This velocity is constant because there is no air resistance. Then, just before the balloon hits the ground, it will have a horizontal velocity of 6 m/s.

b) To calculate the vertical component of the velocity when the balloon hits the ground, let´s use the equation of the vertical component of the velocity:

v1y = v0y + g · t

Since v0y = 0

v1y = -9.8 m/s² · (10 s) = -98 m/s

The magnitude of the vertical velocity of the balloon when it hits the ground is 98 m/s.

4 0

2 years ago

A 0.10 kg piece of copper at an initial temperature of 95°c is dropped into 0.20 kg of water contained in a 0.28 kg aluminum cal

DedPeter [7]

(cp of Copper = 387J / kg times degrees C; cp of Aluminum = 899 J / kg times degrees C; cp of Water = 4186J / kg times degrees C)
 Use the law of conservation of energy and assuming no heat loss to the surroundings, then
Heat given up by copper = heat absorbed by water + heat absorbed by calorimeter
 Working formula is
 Q = heat = MCp(delta T)
 where
 M = mass of the substance
 Cp = specific heat of the substance
 delta T = change in temperature
 Heat given up by copper = 0.10(387)(95 - T)
 Heat absorbed by water = 0.20(4186)(T - 15)
 Heat absorbed by calorimeter = 0.28(899)(T - 15)
 where
 T = final temperature of the system
 Substituting appropriate values,

 0.10(387)(95 - T) = 0.20(4186)(T - 15) + 0.28(899)(T - 15)
 38.7(95 - T) = 1088.92(T - 15)
 3676.50 - 38.7T = 1088.92T - 16333.8
1127.62T = 20010.3
 T = 17.75 C

8 0

2 years ago

A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If

Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, $v_{PA} = 900\ km/h$