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Fudgin [204]
3 years ago
8

As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself.There is no one ar

ound to help, so you attach an ideal pulley tothe box and a massless rope to the ceiling, which you wrap aroundthe pulley. You pull up on the rope to lift the box. Use g or themagnitude of the acceleration due to gravity and neglect frictionforces.
Part A What is the magnitude F of the upward force you must apply to the ropeto start raising the box with constant velocity?

Physics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer: magnitude of applied force is FA = mg + F

Where F is the resultant force downward that the rope moves with

Explanation:

Force downwards F is,

F = FA - T

T is the upwards tension force on the rope

FA is the actual applied force in pulling the rope down.

Therefore, T = FA - F .....equ. (1)

For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)

Therefore, ma = T - mg

T = ma + mg ..... equ. (2)

Equating equ. 1 and 2

T = FA - F = ma + mg

Therefore FA = ma + mg + F

But at constant velocity a = 0

Magnitude of applied force becomes

FA = mg + F

See image below

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If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
Likurg_2 [28]

Answer:

17.8cm

Explanation:

1.3kg --> 4cm

1kg --> 3, 1/13cm

5.8kg --> 18.8cm

6 0
2 years ago
Helpppppppp<br> ill give brainliest
Nina [5.8K]

Answer:

I believe the answer is

B. \: 3 \: red, \: 8 \: yellow, \: 1 \: blue

Explanation:

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4 0
2 years ago
You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje
Over [174]

Answer: a. m = 7.7 kg

              b. V = 435.52 in³

              c. m = 1927 kg

              d. V = 335.37 cm³

              e. m = 3 kg

Explanation: <u>Density</u> is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

\rho=\frac{m}{V}

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

\rho=\frac{m}{V}

m=\rho.V

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

m=\rho.V

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as V=\frac{4}{3}.\pi.r^{3}

Diameter is twice the radius, then r = 4.31 cm.

Volume is

V=\frac{4}{3}.\pi.(4.31)^{3}

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. m=\rho.V

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0
2 years ago
The velocity of sound apparatus is used in an investigation to determine the frequency of an unknown tuning fork. The temperatur
Hitman42 [59]

Answer:

Explanation:

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength.

6 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
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