Ask question

Ask question

All categories

notsponge [240]

3 years ago

8

The wind exerts a force of 425 N to the north on a sailboat, while the water exerts a drag force of 325N in the opposite directi

on. What is the acceleration of the sailboat if the mass is 4100 kg?

1 answer:

Nataly_w [17]3 years ago

8 0

Answer:

its n vv

Explanation:

You might be interested in

What happens in the crushing can experiment?

Dafna1 [17]

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

7 0

3 years ago

A tuning fork has a frequency of 280 Hz and the wavelength of the sound produced is 1.5 meters. Calculate the wave's frequency a

s2008m [1.1K]

Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

3 0

3 years ago

Cell phones do not work everywhere. There are places where your cell phone cannot get a signal

shusha [124]

There are probably no strong signals

5 0

3 years ago

Read 2 more answers

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

3 0

3 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

3 years ago