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Sunny_sXe [5.5K]

3 years ago

8

A person whose mass is m = 70.0 kg m=70.0 kg steps on a mechanical bathroom scale placed on an inclined plane that makes the ang

le α = 22.3 ∘ α=22.3∘ with the horizontal. What is the reading on the scale?

1 answer:

oee [108]3 years ago

8 0

Answer:The reading on the scale is 64.76kg

Explanation:

According to Newton's 3rd law,when a person exerts a force on a plane,the plane simultaneously exerts a force equal in magnitude and opposite.

Given: mass=70kg ,alpha= 22.3°

Mg=mg×cos alpha= 70kg× cos22.3°=

70× 0.9252= 64.76kg

The reading on the scale is 64.76kg

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

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Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads

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