A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?

A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it

Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

In how many seconds will it reach the ground?

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

v = v₀ + at

Substitute known values into the equation.

0 = 200 · sin(30) + (-9.8)t
-200 · sin(30) = -9.8t
t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

2t = 20.40816327

The body will reach the ground in 20.41 seconds.

How far from the point of projection would it strike?

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0

1 year ago

What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s

ICE Princess25 [194]

P = mv
p = 3.5 × 5
p = 17.5 kg .m/s

Hope this helps!

6 0

2 years ago

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

1 year ago

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

so height is 69.68 m

6 0

2 years ago

To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a

Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0

2 years ago