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2 years ago

Which statement is supported by this scenario?

Physics

2 answers:

Temka [501]2 years ago

6 0

Answer:

For Yanni, the speed of the ball is 15 m/s, and for the quarterback, the speed of the ball is 8 m/s.

Explanation:

likoan [24]2 years ago

4 0

Answer:

For Yanni, the speed of the ball is 23 m/s, and for the quarterback, the speed of the ball is 15 m/s.

Explanation:

Right on edge 2021

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Newton discovered that gravity affects the orbits of planets <br><br> True <br> Or<br> False

Svetach [21]

False. Kepler discovered this and Newton was inspired by this.

4 0

3 years ago

Read 2 more answers

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

3 years ago

A printer is connected to a 1.0 m cable. if the magnetic force is 9.1 × x10-5 n, and the magnetic field is 1.3 × 10-4 t, what is

Valentin [98]

The magnetic force on a current-carrying wire due to a magnetic field is given by
$F=ILB$
where
I is the current
L the wire length
B the magnetic field strength

In our problem, L=1.0 m, $F=9.1 \cdot 10^{-5} N$ and $B=1.3 \cdot 10^{-4}T$ , so we can re-arrange the formula to find the current in the wire:
$I= \frac{F}{LB}= \frac{9.1 \cdot 10^{-5} N}{(1.0 m)(1.3 \cdot 10^{-4} T)} =0.7 A$

5 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$