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3 years ago

What would be the force required to accelerate an object that weighs 35 N up at a rate of .60 m/s^2?

Physics

1 answer:

natta225 [31]3 years ago

3 0

Answer: $2.14\ N$

Explanation:

Given

Force required to accelerate the object at the rate of $0.6\ m/s^2$

The weight of the object is $W=35\ N$

Mass of the object can be identified by dividing the weight by acceleration due to gravity

$\Rightarrow m=\dfrac{W}{g}\\\\\Rightarrow m=\dfrac{35}{9.8}\\\\\Rightarrow m=3.57\ kg$

$\therefore \text{Force F=}m\times a\\\Rightarrow F=3.57\times 0.6\\\Rightarrow F=2.14\ N$

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Which part of the neuron receives messages from other neurons?

kow [346]

Answer:

Explanation:

the dendrites is a part of the neuron receives messages from other neurons. Dendrites extend out from the cell body and receive messages from other nerve cells

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2 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

4 years ago

A water pump is a positive displacement-type pump true or false

Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0

4 years ago

A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th

Ipatiy [6.2K]

Answer:

$N_{2}$ =20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by secondary coil I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

$I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}$

putting values we get

$I_{p}=\frac{6}{120\times 0.5}$

$I_{p}= 0.1 A$

to calculate number of turns in secondary

$\frac{N_{2}}{400} =\frac{6}{120}$

therefore, $N_{2}$ =20 turns

5 0

4 years ago

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame

natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

$\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}$

where;

L = length of the bed

$\mu$ = viscosity

U = superficial velocity

$\epsilon$ = void fraction

dp = equivalent spherical diameter of bed material (m)

$\rho$ = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6 -------------- equation (1)

24A + 576B = 24.1 --------------- equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

B = 0.017014

From equation (1)

12A + 144B = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

$A = \frac{9.6 -144(0.017014}{12}$

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0

3 years ago