answer.
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
Maru [420]
1 year ago
13

The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang

le. There is a 1.5 N force of friction to the right. The force of gravity is 58.8 N. What is the normal force acting on the sled? Round the answer to the nearest whole number. N What is acceleration of the sled? Round the answer to the nearest tenth. m/s2
Physics
2 answers:
Troyanec [42]1 year ago
7 0

Answer:

54N and -1.2m/s squared

Explanation:

STatiana [176]1 year ago
6 0

the correct answers are 54N and -1,2m/s^2

You might be interested in
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

5 0
2 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi
Veronika [31]

Answer:

a) The the charges on the inner and outer surfaces of the shell are respectively +q, -q

Explanation:

Under static equilibrium inside a conductor, the total electric field, E = 0

This must be zero so that no charge will be moving since the conductor is in static equilibrium.

Also, since Electric field, E is zero, then flux through the surface will zero.

From Gauss' law, the total charge enclosed is zero.

Given –q as the  charge at the center of the shell, then the opposite charge on inner surfaces  will be +q, so that the total charge enclosed will be zero.

Since the charge is in static equilibrium, then opposite charge will be on the surface, that is –q.

Therefore, the the charges on the inner and outer surfaces of the shell are respectively +q, -q

8 0
2 years ago
A projectile is launched at an angle of 45° from the horizontal and lands 21 s later at the same height from which it was launch
irinina [24]

Answer:

a) initial speed of projectile = 145.5 m/s

b) Maximum altitude = 540 m

c) Range = 2160.6 m

d) r = (1440î + 480j) m

Explanation:

The distance at any time for the projectile is given by the relation - r² = x² + y²

where x = horizontal distance covered covered by the projectile and y = vertical distance coveredby the projectile

Let the initial velocity be u = ?

angle of projection be θ with respect to the horizontal = 45°

u = (uₓî + uᵧj) m/s

T = total time of flight = 21 s

t = any time during the flight of the projectile

a) Total time of flight = 2 uᵧ/g = (2u sin θ)/g

21 = (2u sin 45°)/9.8

u = 145.5 m/s

b) maximum altitude of the projectile = H

H = (u² sin² θ)/2g

H = (145.5² sin² 45°)/(2 × 9.8)

H = 540 m

c) According to projectile motion the maximum horizontal displacement is given by

x = R = uₓT = u cos(θ) T (since uₓ = u cos θ)

R = (145.5 cos 45°) × 21 = 2160.6 m

d) At 14 s,

x = uₓt = u cos(θ) t (since uₓ = u cos θ)

x = (145.5 cos 45°) × 14 = 1440.1 m

y = uᵧ t - 0.5gt² = [u sin(θ)] t - 0.5gt² = (145.5 sin 45°) × 14 - 0.5(9.8)(14) = 480 m

r = (1440î + 480j) m

6 0
2 years ago
In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land
sattari [20]

Answer:

a) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s

a) Impulse

J = final momentum - initial momentum

J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s

Magnitude of impulse

|J| = 219.1 kgm/s

b) Force

F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N

4 0
2 years ago
Other questions:
  • The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light
    9·2 answers
  • A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterbac
    14·1 answer
  • Often what one expects to see influences what is perceived in the surrounding environment. Please select the best answer from th
    5·2 answers
  • A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
    7·1 answer
  • In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
    10·2 answers
  • Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
    8·1 answer
  • a student wants to push a box of books with the mass of 50 kg in 3 m horizontally towards the location of the shelves where the
    11·1 answer
  • If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?
    8·2 answers
  • An astronaut drops a feather from 1.2 m above
    5·1 answer
  • According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!