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2 years ago

The data table depicts an object moving with changing speed. True False

Physics

2 answers:

Nimfa-mama [501]2 years ago

6 0

The answer is true. The table does show an object moving with changing speed.

hoa [83]2 years ago

5 0

Answer:

TRUE

Explanation:

Uniform speed means the object will travel in such a way that it must cover equal distance in equal interval of time.

So after each equal interval of time the distance moved by the object is same then it is uniform speed but if the object travel unequal distance in equal interval of time then it is representing the changing speed.

So here we can see that object covers 1 m distance in first interval of 1 s while in next 1 s interval from t= 1 to t = 2 it moved total distance of d = 5 - 1 = 4 m

In next interval of 1 s from t = 2 to t =3 it moved total distance of d = 25 - 5 = 20 m

So here we can say that speed of the object is changing

So above statement is TRUE

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g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.

docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ( $u^{2}$ *sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ( $u^{2}$ * $sin^{2}$ θ )/2g

the maximum distance also depends upon square of the initial velocity.

7 0

3 years ago

The container contains

monitta

What is the question

7 0

2 years ago

A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find

vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, $a_c$ = v²/r

∴ The centripetal acceleration, $a_c$ = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, $a_c$ = 20·π/9 m/s² ≈ 6.98 m/s²

4 0

2 years ago

Las condiciones iniciales de un gas son 3000 cm3

slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

$PV=nRT$

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

8 0

3 years ago

How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english

Finger [1]

The main formula to be used here is

Force = (mass) x (acceleration).

We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the right side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .

At the same time, both of those will be pulled up by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

Acceleration = (force) / (mass) = 49/10 = 4.9 meters per second²

7 0

3 years ago