Limestone formations in caves are considered what kind of weathering?

At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr

kumpel [21]

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide= $42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to $a = 5.1\ m/s^2$

4 0

3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

2 years ago

Milk has a density of 1.035 g/ml and olive oil has a density of 0.850 g/ml. Which has more mass

marysya [2.9K]

Answer:

Milk.

Explanation:

Density is used to describe how much space an object or substance takes up in relation to the amount of matter in that object or substance (its mass).

Another way to put it is that, density of a substance is the amount of mass of that substance per unit volume. If an object is heavy and compact, it has a high density.

Density = mass/volume

So,

Milk = 1.035 g of milk per 1 ml

Olive oil = 0.85 g of olive oil in 1 ml.

Therefore, milk has a higher mass.

6 0

2 years ago

Read 2 more answers

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

2 years ago