When electric current is flowing in a circuit, if the voltage applied to the circuit is increased, the

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straig

kupik [55]

Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s = 19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

4 0

4 years ago

the video discusses four potential models of how the expansion of the universe changes with time. drag the correct model descrip

lapo4ka [179]

Only the coasting model makes the assumption that the universe's expansion pace never changes.

The current expansion of the cosmos is predicted by all four hypotheses.

Only the recollapsing model foresees the eventual contraction of the universe.

The average mass density of the cosmos is predicted to be exactly the critical density only under the critical model.

The cosmos is only predicted to be expanding more quickly than it did in the distant past by the accelerating model.

Cosmologists use observations of radiation released soon after the Big Bang to determine the expansion rate, often known as the Hubble constant.

Since the unexpected finding that the universe's expansion is speeding up in 1998, cosmologists have incorporated repulsive dark energy into their theory of cosmic development. The Hubble constant's inconsistent values might be resolved by early dark energy, an additional amount of dark energy present in the early cosmos. This early dark energy would have accelerated the universe's expansion by exerting external pressure.

To learn more about Universe expansion please visit-
brainly.com/question/16048709
#SPJ4

8 0

2 years ago

(35 points!!) A block and tackle is used to lift a load weighing 600 lb. The operator lifts the load distance of 1.25 ft by pull

Drupady [299]

Answer:

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Explanation:

Given:

Weight of the load, $W=600$ lb.

Distance moved by the load, $D_{load}=1.25$ ft.

Force applied on the rope, $F = 200$ lb

Distance moved by the force on the rope, $D{in}=4$ ft.

Work done by a force causing a displacement in the direction of force is given as:

$\textrm{Work}=\textrm{Force}\times \textrm{Displacement}$

A.

Here, Input Work is given by the input force and the displacement caused by the input force. So,

$W_{in}=F\times D_{in}\\W_{in}=200\times 4=800\textrm{ ft.lb}$

Therefore, the input work is 800 lb.ft

B.

Output Work is given by the output force and the displacement caused by the output force. So,

$W_{out}=F_{load}\times D_{load}\\W_{out}=600\times 1.25=750\textrm{ ft.lb}$

Therefore, the output work is 750 ft.lb

C.

Efficiency is given as the ratio of Output Work to Input Work expressed as percentage. So,

Efficiency = $\frac{W_{out}}{W_{in}}\times 100=\frac{750}{800}\times 100=93.75\%$

Therefore, efficiency of block and tackle is 93.75 %.

6 0

3 years ago

Read 2 more answers

Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush

Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

6 0

2 years ago

Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.

Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

$distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m$

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

$displacement = +1.0 m-0 m=+1 m$

8 0

3 years ago