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Novay_Z [31]
3 years ago
5

What is the effect on the force of gravity between two objects if the mass of one object doubles?

Physics
1 answer:
Leni [432]3 years ago
6 0
Then the force will also be doubled
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A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

3 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the
inn [45]

Answer:

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force

Explanation:

ig the answer is true

5 0
3 years ago
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
2 years ago
A 1400 kg wrecking ball hangs from a 20-m-long cable. the ball is pulled back until the cable makes an angle of 30.0 ∘ with the
anastassius [24]

From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m. 

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

5 0
3 years ago
Read 2 more answers
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