A machine that uses 200 w of power moves an object a distance of 15 m in 25 sec. Find

PLSSSS HELPPPP PLSSSS 25 POINTS PLS I REALLY NEED THIS AND GET THEM RIGHT SO PLS DONT ANSWER IF U DONT KNOW

denis23 [38]

Answer:

1- C 2-B 3-B - these are ur best answers

Explanation:

4 0

3 years ago

Read 2 more answers

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

$P = I^2R$

make I subject of the formula we have

$I= \sqrt{\frac{P}{R} }$

substitute

$I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps$

8 0

2 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago

A:10i - 2j -4k and B: i +7j - k. Determine |A-B|

maksim [4K]

A - B = (10i - 2j - 4k) - (i + 7j - k)

A - B = 9i - 9j - 3k

|A - B| = √(9² + (-9)² + (-3)²) = √189 = 3√19

8 0

2 years ago

Q1:A large tank is filled with water. The pressure on the base of the fish tank is 4000N/m². The base of the tank is a rectangle

Ymorist [56]

Answer:

F = 36 kN

Explanation:

It is given that,

The pressure on the base of the fish tank is 4000N/m².

The base of the tank is a rectangle measuring 2.0m by 4.5m.

Area of the base of the tank is 9 m²

We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,

$P=\dfrac{F}{A}\\\\F=P\times A\\\\F=4000\times 9\\\\F=36000\ N\\\\F=36\ kN$

So, the force of 36 kN is acting on the base by the base of water.

7 0

3 years ago