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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

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A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

$a_c = \frac{V^2}{r}$

Where,

V = Tangencial velocity

r = radius

With our values we know that

$a_c = \frac{V^2}{r}$

$\frac{V^2}{r} = \frac{1}{10}g$

Therefore solving to find V, we have:

$V = \sqrt{\frac{1}{10}g*r}$

$V = \sqrt{\frac{9.81*200}{10}}$

$V = 14m/s$

For definition we know that the Time to complete are revolution is given by

$t = \frac{Perimeter}{Speed}$

$t = \frac{2\pi R}{V}$

$t = \frac{2\pi * 200}{14}$

$t = 1.5min$

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3 years ago

A hot air balloon contains 85,000 moles of air. To what temperature must the

Oksi-84 [34.3K]

Answer:

B is the best answer for this question

8 0

3 years ago

Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.

julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

5 0

3 years ago

A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light λ is 580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D = 1.8 m.

Width of one fringe = $\frac{\lambda\times D}{d}$

Putting the values we get fringe width

= $\frac{580\times10^{-9}\times1.8}{.000315}$

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be

= 4.86 mm.

3 0

3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago