As a pendulum bob swings back and forth several times, the maximum height it reaches becomes less and less.

A guy wire helping to stabilize a transmitting tower 500 m high makes in angle of 50° with the ground. In a strong wind, the tow

timurjin [86]

They give us the cable tension t = 1500 newton. We assume that the mass of the cable is negligible compared to that of the tower.

We have a force t of 1500 Newton. This force has a vertical component on the y axis and a horizontal component on the x axis.

Of these two components of force, we are especially interested in calculating is the magnitude of the vertical component.

If the angle that it forms with the ground is 50 °, then the vertical component of the force is:

Fy = 1500sin (50)

Fy = 1149 N. <1200 N

The wire will not be loose

6 0

2 years ago

you're riding on a train that's moving at 5 km/h and you roll a ball down the aisle at 2 km/h in the same direction as the train

jekas [21]

Just add the 2 and 5.
7 km/h

3 0

3 years ago

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.

lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

$\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\$

Δtm is time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

$\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr$

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0

3 years ago

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

Assuming that no heat is lost in the surrounding.

We know:

$Q=m.c.\Delta T$

Now the total heat given by the steam to form water at the given conditions:

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

What could be done to change this carbide ion to a neutral carbon atom?

Anarel [89]

The answer is to remove 4 electrons.

3 0

3 years ago