Ask question

Ask question

All categories

Lubov Fominskaja [6]

2 years ago

11

PLEASE HELP!!! i’ll mark brainliest if you’re correct

1 answer:

dedylja [7]2 years ago

7 0

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The net force acting on the block is ~

$\qquad \sf \dashrightarrow \: 10 + 5$

$\qquad \sf \dashrightarrow \: 15 \:N$

So, the Answer in the boxes will be ~

$\boxed{ \sf15} \: \boxed{ \sf N}$

You might be interested in

HELLPPP NOWWA!!!!!!!!!!!!

prisoha [69]

Answer:

b

Explanation:

the gravitational pull also helps with that but

7 0

3 years ago

Read 2 more answers

Name three types of stress that a force can cause inside an object?

Anton [14]

Answer:

Explanation:

Stress is the force applied to a rock and may cause deformation. The three main types of stress are typical of the three types of plate boundaries: compression at convergent boundaries, tension at divergent boundaries, and shear at transform boundaries.

8 0

3 years ago

An iron sphere of radius 0.18m has mass 190 kg. Calculate the density of the iron

Slav-nsk [51]

radius=r=0.18m

Volume:-

$\\ \sf\rightarrowtail \dfrac{4}{3}\pi r^3$

$\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.18)^3$

$\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.005832)$

$\\ \sf\rightarrowtail 0.024m^3$

Mass=190kg

Density:-

$\\ \sf\rightarrowtail \dfrac{Mass}{Volume}$

$\\ \sf\rightarrowtail \dfrac{190}{0.024}$

$\\ \sf\rightarrowtail 7916.67kg/m^3$

8 0

2 years ago

You are holding a string tied to a lamp post 15 m away. You pull the string so that it has a tension of 1.5 N and a mass density

Dvinal [7]

Answer:

Explanation:

Given that

L= 15 m

T= 1.5 N

μ =0.0045 kg/m

f= 4 Hz

We know that

Velocity in string V is given as

$V=\sqrt{\frac{T}{\mu}}$

Now putting the values in the above equation we get

$V=\sqrt{\frac{1.5}{0.0045}}=18.25\ m/s$

We know that

V= f λ

$\lambda =\frac{V}{f}$

$\lambda =\frac{18.25}{4}=4.5 m$

6 0

3 years ago

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$ ....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$ ....(II)

We need to calculate the speed of the object just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$ .

4 0

3 years ago