1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Galina-37 [17]
3 years ago
6

A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. If friction is negligible, what is the resu

ltant force on the sled?
Physics
1 answer:
slega [8]3 years ago
8 0

Answer:

Resultant force = 17.02 N

Explanation:

As we assume the coefficient of friction is negligible, the normal force won't affect the resultant force.

As a result of this, the sine of the angle times the force of gravity on the 10 kg mass is equal to the resultant force, which is, force due to gravity = m × a = 10 kg × 9.8 m/s² = 98 N

98 sin(10)=?

Sin(10)= 0.1736

98 x 0.1736= 17.02 N.

Therefore, the resultant force = 17.02 N

You might be interested in
Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as
Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

4 0
3 years ago
Noble Gases do not readily form compounds because they ___ chemically stable with 8 valence electrons
azamat

Answer : Noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

Explanation :

Noble gases are the chemical elements that are present in group 18 in the periodic table.

The elements are helium, neon, argon, krypton, xenon and radon.

They are chemically most stable except helium due to having the maximum number of 8 valence electrons can hold their outermost shell that means they have a complete octet.

They are rarely reacts with other elements to form compounds by gaining or losing electrons since they are already chemically stable.

Hence, the noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

8 0
3 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
Circular motion formulas
KIM [24]

Answer:

I found this don't know if its any use or not

5 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
Other questions:
  • The international space station orbits the earth at a distance of 6,783 km from the center of the earth it travels at 7.66 km/s.
    10·1 answer
  • Sliding friction is _ than the static friction.
    9·1 answer
  • CAN SOMEONE DO THIS TO ME ITS OK IF YOUR DRAWING IS NOT THAT COOL OR GOOD AND DONT GET A PHOTO FROM SOCIAL MEDIA
    13·1 answer
  • What is SI unit of energy?​
    12·1 answer
  • Draw the distance time graph for a body at rest​
    5·2 answers
  • What is the answer??
    10·1 answer
  • Water vapor is an example of a(n)
    6·1 answer
  • A drag racer, starting from rest, speeds up for 402 m with an acceleration of +17.0 m/s2. A parachute then opens, slowing the ca
    7·1 answer
  • What best describes white light as it travels through a prism? Check all that apply. The light slows down. The light bends in th
    10·1 answer
  • What kingdom are true bacteria in
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!