Ask question

Ask question

All categories

3 years ago

11

An office window has dimensions 3.7 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.972 a

tm, but inside the pressure is held at 1.0 atm. What net force pushes out on the window?

1 answer:

Katyanochek1 [597]3 years ago

8 0

Answer:

22044.3 N

Explanation:

We are given that

Dimension of office window= $3.7m\times 2.1 m$

Area of office window= $l\times b=3.7\times 2.1=7.77 m^2$

Outside air pressure=P1=0.972 atm

Inside pressure,P2=1.0 atm

Pressure=P2-P1=1-0.972=0.028 atm=0.028\times 101325 Pa=2837.1 Pa

1 atm=101325 Pa

Force,F=PA

Using the formula

$F=7.77\times 2837.1=22044.3 N$

Hence, the net force pushes out on the window=22044.3 N

You might be interested in

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

tatyana61 [14]

Answer:

$Q_T=63313.5\ J$

Explanation:

Given:

temperature of skin, $T_s=34^{\circ}C$

initial temperature of steam vapour, $T_v=100^{\circ}C$

latent heat of steam, $L=2256\ J.g^{-1}$

mass of steam, $m=25\ g$

specific heat of water, $c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}$

final temperature, $T_f=34^{\circ}C$

<em>Assuming that no heat is lost in the surrounding.</em>

<u>We know:</u>

$Q=m.c.\Delta T$

<u>Now the total heat given by the steam to form water at the given conditions:</u>

$Q_T=Q_{Lv}+Q_w$ ..............................(1)

where:

$Q_{Lv}=$ latent heat given out by vapour to form water of 100°C

$Q_w=$ heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

$Q_T=m(L+c.\Delta T)$

$Q_T=25(2256+4.19\times 66)$

$Q_T=63313.5\ J$

is the heat transferred to the skin.

4 0

3 years ago

Help! I don’t really know what it’s asking

Misha Larkins [42]

You have to do the math of each and see which one adds up to 66.5

6 0

2 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

If izzy mass is 0.3kg he applide 657.9n force what will be the accelration

Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

F = m x a

Therefore, the acceleration of the body is

a = F / m

= 657.9 N / 0.3 kg

= 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0

3 years ago

Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh

Salsk061 [2.6K]

Answer:

a

Explanation:

8 0

2 years ago