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suter [353]
3 years ago
5

A bag of firewood has a mass of 10.0 kilograms and a weight of

Physics
1 answer:
lorasvet [3.4K]3 years ago
8 0
90 kilograms hope this helps
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A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope
Alex

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

      T - W = m a

The weight

      W = mg

The acceleration can be found by derivatives

     a = dv / dt

     v = 2 t + 0.6 t²

     a = 2 + 0.6 t

We replace

      T - mg = m (2 + 0.6t)

      T = m (g + 2 + 0.6 t)               (1)

Let's look for the time for the speed of 15 m / s

       15 = 2 t + 0.6 t²

       0.6 t² + 2 t - 15 = 0

We solve the second degree equation

        t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

        t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

       t = 3.6 s

Let's calculate from equation 1

       T = 2.00 (9.8 + 2 + 0. 6  3.6)

        T = 27.92 N

4 0
3 years ago
A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff ro
dexar [7]

Required value of initial speed of the bullet be ( 4M/m)√(gL).

Given parameters:

Mass of the bullet =m.

Mass of the bob of the pendulum = M.

speed of the bullet before collision = v

Speed of the bullet after collision = v/2.

Length of the pendulum stiff rod = L.

Let speed transmitted to the pendulum be u.

Using principle of conservation of momentum:

mv = Mu + mv/2

⇒ Mu = mv/2

⇒ u = (m/M)v/2

We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]

To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:

u = √(4gL)

⇒  (m/M)v/2 = √(4gL)

⇒ v =( 4M/m)√(gL).

Hence, minimum required  speed of the bullet be ( 4M/m)√(gL).

Learn more about speed here:

brainly.com/question/28224010

#SPJ1

7 0
1 year ago
A tree is fixed relative to Earth a tree is blank relative to the Sun
RUDIKE [14]

moving,

the sun is a different "observer"

8 0
3 years ago
2 strings both vibrate at exactly 220 Hz. The tension in one of them is then decreased sightly. As a result, 3 beats per second
MAVERICK [17]

Explanation:

Given that,

2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.

The tension in one of them is then decreased sightly, then f_2 will decrese.

Beat frequency, f=f_1-f_2

3=220-f_2

f_2=217\ Hz

So, the new frequency of the string is 217 Hz. Hence, this is the required solution.

3 0
3 years ago
What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward
Yuri [45]

Answer:

q = 2.067 \times 10^{-5}\ C

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

q =\dfrac{mg}{E}

q =\dfrac{0.00141 \times 9.81}{670}

q = 2.067 \times 10^{-5}\ C

Charge of the particle is equal to q = 2.067 \times 10^{-5}\ C

6 0
3 years ago
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