A car starts from rest and accelerates to a speed of 30 m/s in a time of

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

C=1

$\rho=1.21 kg/m^3$

Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

Length,l=23.2 cm= $0.232 m$

Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

$v=9.29 m/s$

4 0

2 years ago

An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)

Alecsey [184]

"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is

   (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

   Weight = (60 kg) x (9.8 m/s²)

      =    588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is

(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

   (0.04) x (588 Newtons) = 23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.

6 0

3 years ago

Which equation is equivalent to Log (26•35) ?

Nady [450]

Answer:

Log (26) + Log (35)

6 0

3 years ago

How do you find the molarity of: 48.2 g of sodium carbonate in 0.500 L of solution

romanna [79]

The answer is why lani

7 0

3 years ago

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s

alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
vₓ₀ = v * cos θ (1)
where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

$x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

$cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
Replacing (7) in (6), we get:

$\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

e)

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
The horizontal component, since it keeps constant, is just v₀x:
v₀ₓ = 13.7 m/s
The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

$v_{fy} = v_{oy} - g*t (9)$

Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:

$v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

$v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$

3 0

2 years ago