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vladimir2022 [97]

2 years ago

11

If you are standing on a weighing scale in an elevator what happens to your weight if the elevator accelerates up or accelerates

down ?
Are you going to weigh less, weigh more or the same?

1 answer:

boyakko [2]2 years ago

7 0

When moving upwards, the normal force from the weighing scale on the person increases so the weight appears to increase. The opposite holds true when the elevator goes downwards.

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An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?

Nady [450]

One of the equations of gravity is this:
${v}^{2} = {u}^{2} + 2gh$
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
${7}^{2} = {0}^{2} + 2 \times 10 \times h$
$49 = 0 + 20h$
$h = \frac{49}{20}$
$h = 2.5m$

6 0

2 years ago

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

PolarNik [594]

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

8 0

2 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

2 years ago

A car accelerates uniformly in a straight line

erik [133]

Answer:

21.59 m/s

Explanation:

recall that one of the equations of motions can be expressed as

v² = u² + 2as

where,

v = final velocity (we are asked to find this)

u = initial velocity = 0m/s (because it says that it starts from rest)

a = acceleration = 3.7m/s²

s = distance travelled = 63 m

simply substitute the known values above into the equation:

v² = u² + 2as

v² = 0² + 2(3.7)(63)

v² = 466.2

v = √466.2

v = 21.59 m/s

3 0

2 years ago

A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at

riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0

2 years ago