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BlackZzzverrR [31]
4 years ago
12

Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

a distance of 2.6 m. What is the mass of either of the trucks?
Physics
1 answer:
HACTEHA [7]4 years ago
3 0

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408*10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M = M_1 = M_2 is the masses of the 2 objects. and R = 2.6m is the distance between them.

F_G = 5.3*10^{-4}N is the attraction force.

5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}

7941265 = \frac{M^2}{2.6^2}

M^2 = 53682948.76

M = \sqrt{53682948.76} \approx 7327 kg or 7.3 tons

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A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s.
gtnhenbr [62]

Answer:

(a) 2400N

(b) Yes

Explanation:

(a) Mass (m) = 1.5kg, initial speed (v) = 4m/s, time (t) = 2.5milliseconds = 2.5/1000 = 0.0025seconds

F = mv/t = (1.5×4)/0.0025 = 6/0.0025 = 2400N

(b) The force would be different because the mass of the two hands and forearms is 3kg (2 × 1.5kg = 3kg)

4 0
3 years ago
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
Salsk061 [2.6K]

Answer:

The fluids speed at a) 0.105\,m^{2}  and b) 0.047\,m^{2} are 2.33\,\frac{m}{s^{2}}  and 5.21\,\frac{m}{s^{2}} respectively

c) Th volume of water the pipe discharges is: 882\,m^{3}  

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

\Delta Q=0(1)

With Q the flux of water, but Q is Av using this on (1) we have:

A_{2}v_{2}-A_{1}v_{1}=0 (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to A_{2}=0.105\,m^{2}, so now we can solve (2) for v_{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}

b) Here we use point 2 as A_{2}=0.047\,m^{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is Q=\frac{V}{t}, so we can write:

A_{1}v_{1}=\frac{V}{t}, solving for V:

V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}

3 0
4 years ago
Which does a reference point provide
mr_godi [17]
Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"
5 0
3 years ago
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
salantis [7]
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force F_f which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
F-F_f=0
so
F=F_f

The frictional force can be rewritten as
F_f = \mu m g
where m=50 kg, g=9.81 m/s^2. Re-arranging, we can solve this equation to find \mu, the coefficient of dynamic friction:
\mu =  \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)}  =0.20
4 0
3 years ago
What is the name of the galaxy in which the earth resides?
Crazy boy [7]
The answer is the Milky Way .
6 0
3 years ago
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