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Angelina_Jolie [31]
3 years ago
6

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

orizontal force must be applied to the box for it to start sliding along the surface?
Physics
1 answer:
MariettaO [177]3 years ago
8 0

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

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What type of soil is found around Pennsylvania riverbanks and consists of particles smaller than sand
PSYCHO15rus [73]

Answer:

Monongahela silt loam

Explanation:

Monongahela silt loam

its obvious

3 0
3 years ago
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surfac
Ostrovityanka [42]

Answer:

Explanation:

Given

acceleration in x direction

a_x=5.10 m/s^2

Final velocity in x direction v_x=3500 m/s

time taken=150 s

let initial velocity in x direction be u_x

v_x=u_x+a_x\times t

3500-5.10\times 150=u_x

u_x=2735 m/s

Now solve for Y direction

v_y=u_y+a_y\times t

4046=u_y+7.3\times 150

u_y=4046-1095=2951 m/s

tan\theta =\frac{u_y}{u-x}=\frac{2951}{2735}

tan\theta =1.078

\theta =47.14 ^{\circ} ACW from x-axis

5 0
3 years ago
What is the speed of an electron traveling 32 cm in 2 ns?
abruzzese [7]

Answer:

Speed, v=1.6\times 10^8\ m/s

Explanation:

Given that,

Distance covered by the electron, d = 32 cm = 0.32 m

Time, t = 2 ns

We need to find the speed of an electron. Speed is equal to distance covered divided by time. So,

v=\dfrac{0.32}{2\times 10^{-9}}\\\\v=1.6\times 10^8\ m/s

So, the speed of the electron is 1.6\times 10^8\ m/s.

6 0
3 years ago
A plane is flying horizontally with speed 237 m/s at a height 1620m above the ground when a package is dropped for the plane. Th
Thepotemich [5.8K]

1. 0.5g*t^2 = 2010 m.

4.9t^2 = 2010.

t = 20.3 s. = Fall time.

D = Xo*t. = 193m/s * 20.3s = 3909 m.

2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s.

3. Vo*t + 0.5g*t^2 = 2010 m.

58*t + 4.9*t^2 = 2010.

4.9t^2 + 58t - 2010 = 0.

Use Quadratic Formula.

t = 15.2 s. = Fall time.

D = 193m/s * 15.2s = 2934 m.

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3 years ago
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Answer:

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5 0
3 years ago
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