If the force exerted by a man is halved and the distance is multiplied by four, by which factor does the work change ?

Use the dimensional analysis and check the correctness of given equation:- PV= nRT (no spam ) plz

gayaneshka [121]

Answer:

PV = nRT

n and R are constants.

Dimensions of Pressure = ML^-1T^-1

Volume dimensions = L³

$(pressure)(volume) = (temperature) \\ (ML {}^{ - 1} T {}^{ - 2} )(L {}^{3} ) = (k) \\ ML {}^{2} T {}^{ - 2} = k$

It is dimensionally inconsistent or incorrect

7 0

3 years ago

Read 2 more answers

How to calculate average breaking force physics

Alexandra [31]

Answer:

Formula: Remove the zero from the speed, multiply the figure by itself and then multiply by 0.4. The figure 0.4 is taken from the fact that the braking distance from 10 km/h in dry road conditions is approximately 0.4 metres.

8 0

3 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

Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d

dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

$W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J$