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marshall27 [118]
4 years ago
12

A change in the gravitational force acting on an object will affect the object’s

Physics
2 answers:
FinnZ [79.3K]4 years ago
5 0
The answer is C, weight
Diano4ka-milaya [45]4 years ago
3 0

c. weight.

Explanation:

I had the exact same question

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The force of friction acting on a sliding crate is 223 N.
lozanna [386]

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

6 0
3 years ago
What is the acceleration of a 5 kg mass pushed by a 10 n force?
pav-90 [236]
<span>2Kg50m/
s2.5m/
s2<span>2m/
s2</span></span>
6 0
3 years ago
Read 2 more answers
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of a uniform electric field of 80.0 kN/C
torisob [31]

Answer:

σ = ±708 nC/m²

Q = ±177 nC

Explanation:

given data

Side of copper plate L = 50 cm

Electric field, E = 80 kN/C

solution

we get here Charge density,σ that is express as

σ = E x ε₀         ....................1

here ε₀ is Permittivity of free space that is 8.85 x 10⁻¹² C²/Nm²

so put value in eq1we get

σ = 80 x 10³ x 8.85 x 10⁻¹²

σ = 708 x 10⁻⁹ C/m²

σ = 708 nC/m²

and

now we get here total change on each faces

Q = σ  A    ...............2

Q = 708 x 10⁻⁹ x (0.50)²

Q = 177 nC

3 0
3 years ago
If a bus travel 200 km in 45 minutes calculate the speed in kilometre per minute​
erma4kov [3.2K]

Answer:

multiply that and divided by 45

6 0
3 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
4 years ago
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