Which equation best summarizes Newton’s 2nd law:

How do I know what kinematic equations to use when solving a question?

olganol [36]

Explanation:

There are 5 kinematic equations, and 5 variables.

Each question will give you 3 variables and ask you to solve for a fourth.

To determine which equation to use, look at which variable is not included in the problem.

For example, if the question does not include time, then you need to use a kinematic equation that does not have t in it. That would be:

v² = v₀² + 2aΔx

Or, if the question does not include the final velocity, then you need a kinematic equation that does not have v in it. That would be:

Δx = v₀ t + ½ at²

5 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

if a body of mass 2kg moving with Velocity of 15m/s collides with a stationary body of same mass then after elastic collision 2n

pochemuha

Answer:

v = 15 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass = 2 [kg]

v = velocity = 15 [m/s]

$P=2*15\\P=30 [kg*m/s]$

Since we know that momentum is conserved, that is, all momentum is transferred to the second body, we can determine the velocity of the second body, since the mass is equal to that of the first body.

$30=2*v\\v = 30/2\\v = 15 [m/s]$

5 0

3 years ago

Your question: "the picture shows a professional diver with a mass of 93 kg from a 25 m high cliff. Earths Gravity is acting on

GuDViN [60]

When we see the words "Which statement ... ", we know right away that there
will be a list of choices, and we're expected to select our answer from that list.
Strangely, the list of answer-choices for this question has been lost.

Similarly, when we see the words "The picture shows ... ", it's hard to fight
the impulse to look around. In the present situation, that's missing too.

If the diver is just standing there, then the reaction force provided by the cliff
against his feet must be exactly equal to his weight. If the vertical forces acting
on the soles of his feet were not balanced, then his feet would be accelerating
vertically.

His weight is (mass) x (gravity) =

(93 kg) x (9.8 m/s²) = 911.4 newtons (about 205 pounds) .

That's also the strength of the upward reaction force provided by the cliff.

6 0

4 years ago

Charges are pushed through an electric circuit by?

Semmy [17]

Answer:

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

Explanation:

7 0

3 years ago

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