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Elodia [21]
3 years ago
13

A net force of 3 N accelerates a mass of 3 kg at the rate of 1 m/s2. The acceleration of a mass of

Physics
1 answer:
Cloud [144]3 years ago
3 0

Answer:

,mkjh ,bkl m,

Explanation:

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What force is the total force felt by an object?
skelet666 [1.2K]

Answer:

net force

Explanation:

Net force felt by an object.

6 0
2 years ago
Read 2 more answers
The position-time equation for a cheetah chasing an antelope is:
pochemuha

Answer:

x = 1.6 + 1.7 t^2      omitting signs

a) at t = 0     x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0     V = 0

c) A = d^2 x / d t^2 = 3.4     (at t = 0  A = 3.4 m/s^2)

d)  x = 1.6 + 1.7 * (4.4)^2 = 34.5    (position at 4.4 sec = 34.5 m)

4 0
2 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

F=\frac{k*q1*q2}{d^{2}}

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0
3 years ago
An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i
Svetllana [295]

Answer:

v \approx 9.312\,\frac{m}{s}

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}

K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}

\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta

\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)

v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}

v \approx 9.312\,\frac{m}{s}

3 0
3 years ago
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