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sleet_krkn [62]

2 years ago

4

In 1995, the average life expectancy in the united states was 75.8 years. the average life expectancy in 2015 was 78.8 years. wh

at is the percentage increase in life expectancy in the last 20 years? choose the closest answer. 3.5%

2 answers:

Lemur [1.5K]2 years ago

7 0

First, we determine the difference between the life expectancy in 2015 and 20 years ago. That is,
difference = 78.8 - 75.8 = 3 years
Then, we divide this difference by the life expectancy in 1995 and multiply the quotient by 100%. That is,
(3/75.8) x 100% = 3.96%
Thus, the increase in life expectancy is approximately 3.96%.

blondinia [14]2 years ago

5 0

Answer: 4%

Explanation: original answer is 3.96%, but the 6 rounds up the 9, and the 9 rounds up to an even 4%.

hope this helps! (:

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A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

2 years ago

Read 2 more answers

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

Initial speed of car 1, $u_1=15i\ m/s$ (east)

Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$900\times 15i +750\times 20j=(900+750)V$

$13500i+15000j=1650V$

$V=(8.18i+9.09j)\ m/s$

The magnitude of speed,

$|V|=\sqrt{8.18^2+9.09^2}$

V = 12.22 m/s

(b) Let $\theta$ is the direction the wreckage move just after the collision. It is given by :

$tan\theta=\dfrac{v_y}{v_x}$

$tan\theta=\dfrac{9.09}{8.18}$

$\theta=48.01^{\circ}$

Hence, this is the required solution.

4 0

2 years ago

Determine the torque applied to the shaft of a car that transmits 225 hp

Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

$T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }$

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

$T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu$

8 0

2 years ago

A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational

Naya [18.7K]

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

4 0

2 years ago

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

oksian1 [2.3K]

Answer:

$\phi _{B} =0.855 T-m^{-2}$

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

$B = \mu _{O}J$

magnetic flux is given as

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

2 years ago