Ask question

Ask question

All categories

2 years 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

2 answers:

Troyanec [42]2 years ago

7 0

Answer:

54N and -1.2m/s squared

Explanation:

STatiana [176]2 years ago

6 0

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

You might be interested in

Bricks and insulation are used to construct the walls of a house. The

ira [324]

Answer:

$\dot Q=350.438\ W$

Explanation:

Given:

<u>the thermal resistance in the form of </u>

$R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}$

$R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}$

where:

$x_1\ \&\ x_2$ are the thickness of the respective bricks

$k_1\ \&\ k_2$ are the respective coefficient of conductivity

temperature inside the house, $T_h=24\ ^{\circ}C$

temperature outside the house, $\ T_c=10^{\circ}C$

area of the wall, $A=20\ m^2$

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

<u>Using Fourier's law:</u>

$\dot Q=k.A.\frac{dT}{x}$

$\dot Q={dT}\div {\frac{x}{k.A} }$

in series the resistances get add up

$\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )$

$\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )$

$\dot Q=350.438\ W$

7 0

2 years ago

The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

4 0

2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg </em>

<em>x1 = -0.75 m </em>

<em>x2 = -0.2 m </em>

<em>g = 9.8 m/s^2 </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

2 years ago

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

<span>The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>

3 0

2 years ago

In a photoelectric effect experiment, electromagnetic radiation containing a finite distribution of wavelengths shines on a meta

Sonja [21]

Answer with Explanation:

a.Intensity of radiation is directly proportional to the frequency of radiation

When the intensity of radiation increases then the frequency of radiation increases and therefore, the number of photo-electrons emitted by the metal increases.

b.When all of the wavelength in the radiation are increased by the same amount

We know that

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

Frequency is inversely proportional to the wavelength.

Therefore, the frequency decrease .

When the frequency decreases then the number of photo-electrons emitted by the metal decrease.

c.When the work function of the metal is increased then the gain of kinetic energy decreases .

When energy decreases then the number of photo-electrons emitted by the metal decreases.

3 0

2 years ago