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aliya0001 [1]
3 years ago
6

A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivo

t point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm

Physics
1 answer:
mina [271]3 years ago
8 0

Answer: The magnitude of torque is 38.7Nm

Explanation: Please see the attachment below

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Aleksandr-060686 [28]
I think that the answer is c but I’m not sure
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Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t
Andre45 [30]

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

g=\frac{GM}{R^2}=9.70 m/s^2

where G is the gravitational constant

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R=6378km = 6.378 \cdot 10^6 m is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg

The Earth's volume is (approximating the Earth to a perfect sphere)

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3

So, the average density of the Earth is

\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3

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Answer:

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