a person runs 27.0km west then turns around and runs 13.0km east what's the distance and displacement ?

Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide deta

Paladinen [302]

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

6 0

2 years ago

Soil formation involves both

Softa [21]

The process of soil formation involves the following steps: Accumulation of materials, leaching and losses (because of weather factors particles are leached and eroded away or taken up from the soil by plants), Transformation and illluviation (the chemical weathering<span> of silt, sand, and the formation of clay minerals as well as the change of organic materials into decay resistant organic matter), p</span>odsolisation and translocations (strong acidic solutions breakdown the clay minerals). From the given options soil formation involves both <span>physical and chemical weathering of rocks. Correct answer:D</span>

6 0

2 years ago

When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity

Readme [11.4K]

Answer:

Explanation:

In order to measure the coefficient of friction , we apply external force to move the body . When external force comes in motion , we adjust the external force so that it moves with zero acceleration or uniform velocity . In this case external force becomes equal to kinetic frictional force and then net force becomes zero because

net force = mass x acceleration = m x 0 = 0

Now frictional force = μ mg where μ is coefficient of kinetic friction

so F = μ mg where F is external force applied

μ = F / mg

Hence , to make external force equal to frictional force , it is necessary to make acceleration of body zero .

4 0

2 years ago

The radius of curvature of a spherical mirror is 20cm.What is its focal length?

hichkok12 [17]

Answer:

$\boxed{\sf Focal \ length = 10 \ cm}$

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

$\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}$

Substituting value of R in the equation:

$\sf \implies f = \frac{20}{2}$

$\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}$

$\sf \implies f = 10 \: cm$

5 0

3 years ago

Read 2 more answers

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago