A pressure wave is what type of wave

. Egbert is in a rowboat four miles from the nearest point on Egbert a straight shoreline. He wishes to reach a house 12 miles f

swat32

Answer:

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

Explanation:

Given:

Distance from the shore, $s=4\ mi$

distance of house from the shore, $h=12\ mi$

speed of rowing, $v_r=3\ mi.hr^{-1}$

speed of walking, $v_w=4\ mi.hr^{-1}$

For the least amount of time to reach the house one must row at the nearest point on the shore and then walk from there.

Now the time taken to reach the shore:

$t_s=\frac{s}{v_s}$

$t_s=\frac{4}{3}\ mi.hr^{-1}$

Time taken in walking to house from the shore:

$t_h=\frac{h}{v_w}$

$t_h=\frac{12}{4}$

$t_h=3\ mi.hr^{-1}$

Therefore total time taken:

$t=t_h+t_s$

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

6 0

3 years ago

an 1800-w toaster, a 1400-w electric frying pan, and a 75-w lamp are plugged into the same outlet in a 15-a, 120-v circuit. (the

Gnom [1K]

The current drawn by the electric frying pan is 27.295 A.

Power of toaster, P1 = 1800 W

Power of electric frying pan, P2 = 1400 W

Power of lamp, P3 = 75 W

Three devices are in parallel combination. So their Voltage difference is the same.

∆V = 120 Volt

Current is drawn by frying pan, I1 = P1/∆V = 1800/120 = 15 A

Current drawn by an electric frying pan, I2= 1400/120 = 11.67 A

Current drawn by frying pan, I3 = P3/120 = 75/120 = 0.625 A

Total current frying pan= (15 + 11.67 + 0.625) = 27.295 A

Electric current refers to the float of electricity in an electronic circuit, and to the amount of strength flowing via a circuit. it is measured in amperes (A). the bigger the price in amperes, the greater electricity is flowing inside the circuit.

The current is the charge of glide of price across a cross-segment every day to the direction of the float of the current. The S.I. unit of modern is coulomb consistent with 2d which is called ampere and denoted by means of A.

Learn more about current here:-brainly.com/question/24858512

#SPJ4

8 0

2 years ago

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

Darya [45]

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

a.

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

b.

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

6 0

3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

4 0

3 years ago

What are two types of electric currents?

s2008m [1.1K]

Answer:

There are two different types of current in widespread use today. They are direct current, abbreviated DC, and alternating current, abbreviated AC. In a direct current, the electrons flow in one direction.

6 0

3 years ago

A pressure wave is what type of wave​

A pressure wave is what type of wave