175 in expanded form

The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m

butalik [34]

$\sf\underline{Solution:}$

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

$\sf{Now,according \: to \:the\:question:}$

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

☆ Substitute the required values in the given formula-

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0

2 years ago

When is there the least amount of heat transfer within a liquid

Annette [7]

In the beginning, the melting stage they call it.

6 0

2 years ago

Read 2 more answers

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s

alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
vₓ₀ = v * cos θ (1)
where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

$x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

$cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
Replacing (7) in (6), we get:

$\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

e)

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
The horizontal component, since it keeps constant, is just v₀x:
v₀ₓ = 13.7 m/s
The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

$v_{fy} = v_{oy} - g*t (9)$

Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:

$v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

$v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$

3 0

2 years ago

Which object has the greatest momentum...and why

damaskus [11]

The formula for momentum is p=mv where p is the momentum (kgms-1), m is the mass (kg) and v is the velocity (ms-1). So, to work out the momentum, we just multiply these numbers together and work out which is the largest.

1) 12kgms-1
2) 10kgms-1
3) 27kgms-1
4) 16kgms-1

Therefore the object with the greatest momentum is 3 - a 9kg mass moving at 3m/s

7 0

2 years ago

Read 2 more answers

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$