HELP!!! how does gravity affect how objects fall to the ground

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

1. false

2. true

I hope this helps ^-^

Answer:

e. the number of active motor units increases.

Explanation:

There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.

Answer:

1.61 N/kg

Explanation:

Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²

The required gravitational field strength

= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²

= 1.61194103 N/kg

= 1.61 N/kg (corr. to 3 sig. fig.)

An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time

The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.

As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.

The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum

T = 2π√(L/g)

As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time

Thus, if a  pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder

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The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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Answer:

0.002 m/s

Explanation:

27.13(60) = 1,627.8 seconds

3.12/1,627.8 = 0.00191 ≈ 0.002 = s

Are you sure you're not looking for cm/s?

Answer:

The movement of Sun and moon

Explanation:

When the sun rise.it is am and when it sets .it is pm.

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

ks= 133.2 N/m

Explanation:

       [tex]\Delta U + \Delta K = 0 (1)[/tex]

       [tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

How many mL is an espresso?

One shot of espresso is generally about 30–50 ml (1–1.75 oz), and contains about 63 mg of caffeine (3). Important point: The “golden ratio” for espresso is this: a single shot is 30 to 44 mL (1 to 1.5 ounces) of water and 7 grams of coffee

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

In metals there are free electrons at normal temperature so when we increase temperature it resistivity gets increases,so conductivity decreases,while in semiconductor the electrons are not free so when we increase the temperature the covalent bonds begin to break and the electron becomes free so conductivity get.

Explanation:

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the axial component of the anchoring force required to hold the contraction in place is 365.6 lb

Explanation:

Given the data in the question and as illustrated in the image below;

first we calculate the area at section 1

A₁ = (πD²)/4

we substitute

A₁ = (π(3 in)²)/4

A₁ = 7.06858 in²

we know that; 1 ft = 12 in

A₁ = ( 7.06858 / (12²) ) ft²

A₁ = ( 7.06858 / 144 ) ft²

A₁ = 0.0491 ft²

now, we write the elation for area at section 2

A₂ = πd²/4

here, d is the diameter at section 2

next, we use the conservation of mass equation between the two section;

m" = pV₁A₁ = pV₂A₂

we calculate the mass flow rate;

m" = pV₁A₁

= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²

=  2.8576 slug/s

Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2

-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂

m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂

F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )

we substitute

F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s

F[tex]_A[/tex] =  565.632 - 0 - 200.032

F[tex]_A[/tex] = 565.632 - 200.032

F[tex]_A[/tex] = 365.6 lb

Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

Answer:

Molecular scale. The story begins a long time ago  

when the idea that molecules are in constant motion  

was first discovered. Part of the evidence that you can  

see in everyday life was discovered by Robert Brown  

about 150 years ago when he used a microscope to  

watch how tiny dust particles move.

So how fast do molecules move? It all depends upon  

the molecule and its state: molecules in a solid state  

move slower than in a liquid state, and much slower  

than gas molecules. One estimate puts gas molecules  

in the range of 1,100 mph at room temperature. Cool  

them down to almost absolute zero and they slow  

down to less than 0.1 mph (slower than the average  

couch potato). The fact that they are always moving  

makes it a challenge to see molecules and make stuff  

out of them, but it’s a challenge that scientists  

work hard to figure out.

Explanation:

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

Answer: The daughter is named Susie.

Explanation: LIL SUSIE!!!

                      HUH? DIDN'T UNDERSTAND THE QUESTION!

                                        HAVE A GREAT DAY!!!!!

Answer:6/3 Li

Explanation:

I’m not sure what the person under me is talking about but yeah

Answer:

B. in the direction of the force

Explanation:

Sana nakatulong