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3 years ago

What is the weight of a 24.52kg Television dropped on pluto(acceleration of 0.59m/s/s)?

Physics

2 answers:

jolli1 [7]3 years ago

6 0

W=ma=24.52*0.59 = ...

ella [17]3 years ago

6 0

Answer:

14.47 Newtons

Explanation:

Hello

The weight of an object is the force of gravity on the object and can be defined as the product of mass by the acceleration of gravity, w = mg

for example

a person with mass of 80 kilograms on earth where the acceleration of gravity is 9.8 m/s2 weigh.

w=mg

W=80kg*9.8 m/s2

W=784 Newtons

Now, let see the data we have about the question

$W=mg\\W=24.52 kg* 0.59 \frac{m}{s^{2} } \\W=14.47\ Newtons$

the weight is 14.47 N

Have a great day

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2 years ago

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11 months ago

Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a

Nonamiya [84]

Answer:

$1486.5\frac{Btu}{s}$

Explanation:

The inlet specific volume of air is given by:

$v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \ \ \ \ \ \...i$

The mass flow rates is expressed as:

$\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}$

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$E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}$

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2 years ago

An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.

Minchanka [31]

Answer:

Fₓ = 0, $F_{y}$ = 0 and $F_{z}$ = - 3.115 10⁻¹⁵ N

Explanation:

The magnetic force given by the expression

F = q v xB

the bold are vectors, the easiest analytical way to determine this force in solving the determinant

$F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}$

F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

F =i^0 + j^0 - k^ 3.115 10⁻¹⁵ N

Fₓ = 0

$F_{y}$ = 0

$F_{z}$ = - 3.115 10⁻¹⁵ N

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3 years ago

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Brrunno [24]

Explanation:

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For a negative charge alone (image 2), the lines come from infinity to the centre, and point towards the particle (i.e. lines appear from the edge of the picture).

Let's see what happens if we have two charges together:

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Negative and positive charges (image 5): They are different charges, so the force between them is attractive. This causes the field lines from both to join. They go out of the positive and come into the negative particle.

Image 6:

The lines are passing through infinite points of the space. If we choose a certain point and measure the electric field, we can see to which direction the electric field points. This is the direction of the electric field vector. It does not matter which point we choose; the electric field vector touches the field line only at this point, which means it is tangent to the field line.

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