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2 years ago

A center-seeking force related to acceleration is _______ force.

1 answer:

4 0

<span>A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.</span>

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If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$

$\alpha = - 2.8 rad/s^2$

$t = 1.5 s$

now from kinematics we have

$\omega_f = \omega_i + \alpha t$

now from above all values we have

$\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)$

$\omega_f = -8.4 + (-4.2)$

$\omega_f = -12.6 rad/s$

so final angular speed is -12.6 rad/s

6 0

2 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

2 years ago

What type of data allows geologists to know what rock layers lie underneath Minnesota?

mrs_skeptik [129]

The answer is well log data, it is a detailed log of information taken from a borehole which geologist used to study geological formations of the earth's layer taken from samples returned from the borehole which was dugged.

5 0

2 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

6 0

2 years ago

An unknown substance from planet X has a density of 10 g/mL. It occupies a volume of 80 mL. What is the mass of this unknown sub

asambeis [7]

Answer:

800 mL

Explanation:

D*V=M

You pick out the numbers as well as what it is they represent from the word problem/explanation, then from there plug them in to the equations. Once you do that, you get your product and have the answer.

10*80= 800

5 0

2 years ago