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OlgaM077 [116]
4 years ago
5

Activity level for the following samples?

Physics
1 answer:
Oxana [17]4 years ago
8 0
<span>- 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 S 50 R 51 1st Pass Pages  of Inertia 94 4-2 Force and Mass 95 4-3 Newton's Second Law 97 4-4 The Force The decay constant is related to the half-life by (Equation M-70).</span><span>
</span>
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¿Cuál de las siguientes no es un tipo de fuerza de roce
Degger [83]
Enertia es el answer de tu question
8 0
3 years ago
2. Sally put her mug of coffee on the table. The mug has a mass of 1 kg. if the pressure
egoroff_w [7]

We have,

  • The mass of sally's mug is 1 kg
  • The pressure appliedby the mug is 1100 pascal.

We know that,

  • Pressure = Force/Area

As, we already have the value of pressure, let's calculate that of force now;

  • F = ma
  • F = 1 × 9.8 { Acceleration due to gravity, let's round off it to 10}
  • F = 10 N

Just put all the values in the formula now;

  • P = F/A
  • 1100 = 10 / A
  • 1100/10 = A
  • 110 m² = A

As, it is already mention that we need to find the radius of the mug, it is obviously a circular base.

We know that,

  • Surface area = Circumference

So, let's solve it;

  • Circumference = 2πr
  • 110 = 2 × 22/7 × r
  • 110 × 7/2 × 22 = r
  • 5 × 7 = r
  • 35 cm = r

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7 0
3 years ago
Is energy transformation from potential to kinetic 100%?
Arada [10]
Yes. take a bow for instance. while pulling back the string you have potential energy. when you let the string go and the arrow flies towards your target the string is filled with kinetic energy.
8 0
3 years ago
A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component o
Katen [24]

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, a=49\ m/s^2

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

F=75\ kg\times 49\ m/s^2

F = 3675 N

Ratio, R=\dfrac{F}{W}

R=\dfrac{3675}{75\times 9.8}=5

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

F'=\sqrt{F^2+W^2}

F'=\sqrt{(3675)^2+(75\times 9.8)^2}

F' = 3747.7 N

The direction of force is calculated as :

\theta=tan^{-1}(\dfrac{W}{F})

\theta=tan^{-1}(\dfrac{1}{5})

\theta=11.3^{\circ}

Hence, this is the required solution.

8 0
4 years ago
A satellite is spinning at 0.01 rev/s. The moment of inertia of the satellite about the spin axis is 2000 kg · m2. Paired thrust
enyata [817]

Answer:

Explanation:

Angular velocity of satellite

= 2π x .01

= .02 π rad /s

Initial angular momentum

Moment of inertia x angular velocity

= 2000 x .02 π

= 125.6 unit

Linear impulse produced by each thruster

= 15 N.s

Angular impulse

= 15 x 1.5 = 22.5 unit

Total angular impulse in 30 pulses

= 22.5 x 2 x 30

1350

This angular impulse will add total angular momentum of

1350 unit

So total angular momentum after 30 pulses

= 1350 + 22.5

= 1372.5 unit

So  final angular  velocity

= final  angular momentum / moment of inertia

= 1372.5 / 2000

=   0 .686 rad /s

3 0
3 years ago
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