Ask question

Ask question

All categories

3 years ago

7

Eric has a mass of 80 kgkg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s2m/s2. What is the

approximate reading on the scale?

1 answer:

kolezko [41]3 years ago

5 0

Answer:

The answer will be 936 N.

Explanation:

Given that

m = 80 kg

Acceleration of the elevator , a= 1.7 m/s² ( upward)

The gravity force on the mass = m g

The reading on the scale = F N

Now by applying the Newton's second law

F - m g = ma

F= m g + m a

F= m ( g +a )

F= 80 ( 10 + 1.7 ) N ( take g= 10 m/s²)

F=80 x 11.7 N

F= 936 N

Therefore the reading on the scale will be 936 N.

The answer will be 936 N.

You might be interested in

Carol drops a stone in a mine shaft 122.5 metres deep.How

Verizon [17]

Answer:

T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken = ?

time taken by the stone to reach at the bottom

using equation of motion

$s = u t + \dfrac{1}{2}gt^2$

initial speed , u = 0 m/s

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 122.5}{9.8}}$

t = 5 s

time taken by the sound to travel

d =v x t

$t = \dfrac{d}{v}$

$t = \dfrac{122.5}{340}$

t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

T = 5.36 s

7 0

3 years ago

Which of the following types of rocks are likely to be formed due to heat and pressure? Sedimentary rock only Metamorphic rock o

KiRa [710]

It is metamorphic rock only

6 0

3 years ago

Read 2 more answers

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0

3 years ago

HELP ASAP! GIVING BRAINLIEST!!!<br><br> I need the answers to 1,2,3!!

artcher [175]

1) 72/4 = 18 years

2) 72/10 = 7.2 years

3) 72/13 = 5.54 years

Explanation: With the rule of 72, you simply take 72 and divide it by whatever the rate of return (or return on investment) is.

Let me know if you have any questions.

7 0

3 years ago

Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from

kodGreya [7K]

Answer:

$ax = 2.60m/s^{2}$ , $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

Where $(vx)f^{2}$ is the final velocity, $(vx)i^{2}$ is the initial velocity, $ax$ is the acceleration and $\Lambda x$ is the distance traveled.

Equation (1) can be rewritten in terms of ax:

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$ (2)

Since the plane starts from rest, its initial velocity will be zero ( $(vx) = 0$ ):

Replacing the values given in equation 2, it is gotten:

$ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}$

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

So, The acceleration of the plane is $2.60m/s^{2}$

Now that the acceleration is known, the next equation can be used to find out the time:

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

5 0

3 years ago