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Tems11 [23]
4 years ago
5

What do you call an increase in the magnitude of velocity

Physics
1 answer:
Vlad1618 [11]4 years ago
4 0

Any change in the magnitude or direction of velocity is "acceleration".

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Objects 1 and 2 attract each other with a gravitational force of 45 units. If the mass of Object 1 is doubled, then the new grav
Novosadov [1.4K]

Explanation:

Fgravity = G*(mass1*mass2)/D²

so, if you double one of the masses, what does that do to our equation ?

Fgravitynew = G*(2*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity

so, the correct answer will be 2×45 = 90 units.

4 0
3 years ago
Read 2 more answers
What real-world examples show no work begin done? Can you think of examples other than resisting the force of gravity?
irga5000 [103]
Oh my gosh !  Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.

Example:

-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity

6 0
3 years ago
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Una mujer esta segando su prado con segadora eléctrica cuya cuchilla gira a 1500rpm. La mujer encuentra una zona de hierba alta
maks197457 [2]

Answer:

α= 15.4[rad/s^2], 500 [rev/min]

Explanation:

La aceleración angular constante esta definida por la siguiente ecuación:

w=w_{0} + \alpha *t\\Donde:\\w=2000 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=209.4[\frac{rad}{s} ]\\w_{0} =1500 [\frac{rev}{min}]*[\frac{1min}{60s} ] *[\frac{2\pi rad}{1 rev} ]=157.1[\frac{rad}{s} ]\\t=3.4[s]\\reemplazando:\\209.4-157.1=\alpha *(3.4)\\\\alpha =15.4[\frac{rad}{s^{2} } ]

Para poder llegar a las 2000 [revoluciones /min] estando en 1500 [rev/min], la cuchilla tuvo que incrementar su numero de revoluciones a 500 [rev/min]. Mientras aceleraba.

3 0
3 years ago
What is the total number of natural element ?​
Kitty [74]

Answer:

5 element's

<h3>hope it's help </h3>
8 0
3 years ago
A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision w
patriot [66]

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

4 0
3 years ago
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