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3 years ago

Which of the following forces can change velocity?

Physics

2 answers:

Arte-miy333 [17]3 years ago

5 0

Answer:

Option A and Option D are correct.

Explanation:

We have the Newton's second law, that is rate of change of momentum is force.

$\frac{dP}{dt}=F\\\\\frac{mv-mu}{dt}=F$

So if there is a force mv - mu ≠ 0

v - u ≠ 0

v ≠ u

Final velocity ≠ Initial velocity.

a) Option A is correct since there is a force

b) Option B is wrong since total force is zero.

c) Option C is wrong since total force is zero.

d) Option D is correct since there is an effective force

Option A and Option D are correct.

andrezito [222]3 years ago

4 0

Unequal opposite forces.
(They're unbalanced.)

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An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

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3 years ago

When did agriculture first began?

timofeeve [1]

Answer:

Humans invented agriculture between 7,000 and 10,000 years ago, during the Neolithic era, or the New Stone Age. There were eight Neolithic crops: emmer wheat, einkorn wheat, peas, lentils, bitter vetch, hulled barley, chickpeas, and flax.

5 0

3 years ago

Read 2 more answers

A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together

Lena [83]

Answer:

a. They will be tie

b. Win the wood cylinder

Explanation:

The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.

$I=\frac{1}{2}*m*r^2$

$a=\frac{g*sin(\beta)}{1+I_{com}/m*r^2}$

So both will be tie

$a_{brass}=\frac{g*sin(\beta)}{1+I_{brass}/m*r^2}=a_{wood}=\frac{g*sin(\beta)}{1+I_{wood}/m*r^2}$

The acceleration of the wood cylinder is larger than the acceleration of the brass cylinder so the cylinder of wood will reach the bottom first

$a_{brass}$

So the wood win the race

6 0

3 years ago

Please help!

andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0

3 years ago

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has

a_sh-v [17]

Answer:

h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

Em₀ = U = m g h

final point. Lower, just before the crash

Em_f = K = ½ m $v_{b}^2$

energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2m 0 + m v_b

final instant. Right after the crash

p_f = (2m + m) v

the moment is preserved

p₀ = p_f

m v_b = 3m v

v = v_b / 3

v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

Em₀ = K = ½ 3m v²

Final point. Higher

Em_f = U = (3m) g h'

Em₀ = Em_f

½ 3m v² = 3m g h’

we substitute

h’= $\frac{v^2}{2g}$

h’ = $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

h’ = 1/9 h

6 0

3 years ago