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Veseljchak [2.6K]

3 years ago

9

What is the relationship between the force of gravity between two objects in regard to their mass and the distance between them?

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

hi

Explanation:

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Conveyor belts are often used to move packages around warehouses. The conveyor shown below moves packages at a steady 4.0 m/s. A

Alisiya [41]

Answer:

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Explanation:

4 0

3 years ago

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

Masteriza [31]

If you can write this in English I can help

8 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe

liubo4ka [24]

Answer:

ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

$I_1=\dfrac{M+m}{2}r^2$

$I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2$

$I_1=0.75\ kg.m^2$

Final mass moment of inertia

$I_2=\dfrac{M}{2}r^2$

$I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2$

$I_2=0.468\ kg.m^2$

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

$\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s$

ω₂=1.20

7 0

3 years ago

A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2

Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

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

$a=\dfrac{10\ m/s}{1.28\ s}$

$a=7.81\ m/s^2$

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, $a=\dfrac{36}{0.000355}$

$a=101408.45\ km/h^2$

Hence, this is the required solution.

3 0

3 years ago