Motion maps for two objects, Y & Z, are shown.

A boy and his skateboard have a combined mass of 65 kg what is the speed of the boy and skateboard if they have a momentum of 27

kvasek [131]

Let m = 65 kg, the mass of the boy and skateboard.
Let v = speed of the boy and skateboard.

The momentum is
P = mv = (65 kg)*(v m/s) = 275 (kg-m)/s²
v = 275/65 = 4.23 m/s

Answer: 4.23 m/s

3 0

3 years ago

Two objects of equal mass are a fixed distance apart. If the mass of each object could be tripled, the gravitational force betwe

antiseptic1488 [7]

gravitational force between two objects is given as

F = G m₁ m₂/r²

where m₁ = mass of first object , m₂ = mass of second object , r = distance between the two objects .

Initial case :

m₁ = m₂ = m

gravitational force between the objects is given as

F = G m²/r²

Final Case :

m₁ = m₂ = 3 m

new gravitational force between the objects is given as

F' = G (3m)²/r²

F' = 9 G m²/r²

F' = 9 F

hence the gravitational force between the two objects becomes 9 times.

5 0

3 years ago

Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling

quester [9]

Answer:

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

Explanation:

The chew toy is at equilibrium and experimenting three forces from three distinct dogs. The Free Body Diagram depicting the system is attached below. By Newton's Laws we construct the following equations of equilibrium: (Sp is for Spot, F is for Fido and St is for Steinberg) All forces and angles are measured in newtons and sexagesimal degrees, respectively:

$\Sigma F_{x} = F_{F}\cdot \cos \theta_{F} + F_{St,x} = 0$ (1)

$\Sigma F_{y} = F_{F}\cdot \sin \theta_{F}-F_{Sp}+F_{St,y} = 0$ (2)

If we know that $F_{F} = 20\,N$ , $F_{Sp} = 30\,N$ and $\theta_{F} = 63^{\circ}$ , then the components of the force done by Steinberg on the chewing toy is:

$F_{St,x} = -F_{F}\cdot \cos \theta_{F}$

$F_{St,x} = -(20\,N)\cdot \cos 63^{\circ}$

$F_{St, x} = -9.080\,N$

$F_{St,y} = F_{Sp}-F_{F}\cdot \sin \theta_{F}$

$F_{St,y} = 30\,N-(20\,N)\cdot \sin 63^{\circ}$

$F_{St, y} = 12.180\,N$

The magnitud of the force is determined by Pythagorean Theorem:

$F_{St} = \sqrt{F_{St,x}^{2}+F_{St,y}^{2}}$

$F_{St} =\sqrt{(-9.080\,N)^{2}+(12.180\,N)^{2}}$

$F_{St} \approx 15.192\,N$

Since the direction of this force is in the 3rd Quadrant on Cartesian plane, we determine the direction of the force with respect to the eastern semiaxis:

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{F_{St,y}}{F_{St,x}}\right)$