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3 years ago

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Gravitational force is a(n)? A) attractive force B) repulsive force C) both an attractive and repulsive force D) neither an attr

active or repulsive force

1 answer:

aleksley [76]3 years ago

8 0

Answer: In the case of gravity, mediated by spin 2 particles, charge is mass, which is always positive. Thus, q1q2 is always greater than zero, and gravity is always attractive. For spin 0 force mediators, however, there is no restriction on the charges and you can very well have repulsive forces.

Explanation: C) both an attractive and repulsive force is the answer

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

MaRussiya [10]

It would have to be c because it is a chemical change. your welcome!!!!!

3 0

3 years ago

Read 2 more answers

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A

ki77a [65]

Answer:

point b

Explanation:

7 0

3 years ago

lutik1710 [3]

Answer:

Thats her fault.........................b

Explanation:

6 0

3 years ago