Suppose two carts, one twice as massive as the other, fly apart when the

180 J of work is done when lifting a box up to a shelf that is 3 m high. What is the mass of the box?

ale4655 [162]

Explanation:

Work done= MGH

180=m×10×3

180/30=m

m=6kg

4 0

2 years ago

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

8 0

2 years ago

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

3 years ago

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exe

PtichkaEL [24]

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

4 0

3 years ago

A collapsible plastic bag (figure below) contains a glucose solution. If the average gauge pressure in the vein is 1.29 A collap

expeople1 [14]

Answer:

0.125 m

Explanation:

Pressure in fluids is given as the product of density, height and acceleration due to gravity and expressed as

P=hdg

Where h is the height, d is density, g is acceleration due to gravity and P is pressure.

Making h the subject of formula then

h=P/dg

Given specific gravity of a substance, its density is equal to specific gravity multiplied by density of water. Taking density of pure water as 1000 kg/m³ then the density of reference fluid will be 1.05*1000=1050 kg/m³

Substituting pressure with 1.29*10³ pa as given then taking g as 9.81 m/s² then

H=1.29*10³÷(9.81*1050)=0.1252366389981068880151448958788408329692m

Rounded off, the height is approximately 0.125 m

8 0

3 years ago