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2 years ago

11

A 51.1-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.622 and 0.368,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

1 answer:

Charra [1.4K]2 years ago

4 0

Answer:

(a) The force must be greater than 311.4851 N to just start the crate moving

(b) The 184.287 N force needed to to slide the crate at constant speed

Explanation:

We have given that mass of the crate m = 51.1 kg

Coefficient of static friction $\mu _s=0.622$ and coefficient of kinetic friction $\mu _k=0.368$

Acceleration due to gravity $g=9.8m/sec^2$

(A) Static frictional force is given by $F_S=\mu _Smg=0.622\times 9.8\times 51.1=311.48516N$

So the force must be greater than 311.4851 N to just start the crate moving

(b) For sliding the crate across the dock at a constant speed force must be equal to kinetic friction force

Kinetic friction force is given by $F_k=\mu _kmg=0.368\times 9.8\times 51.1=184.287N$

So the 184.287 N force needed to to slide the crate at constant speed

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Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

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force would a proton experience = ?

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we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

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Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

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2 years ago

Superman is flying 54.5 m/s when he sees

Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$ , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$ , $v_{0}$ and $t$ , we have to find the aceleration $a$ in order to find $v_{f}$ .

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$ , getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$ .

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$ .

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$

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2 years ago

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

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For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
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M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

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2 years ago

At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g

Strike441 [17]

Answer:

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Explanation:

Given that

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Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

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boyakko [2]

Explanation:

It is given that,

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Then the truck moves, $d_2=2.45\ km$ (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

$d=\sqrt{d_1^2+d_2^2}$

$d=\sqrt{3.2^2+2.45^2}$

d = 4.03 km

Let $\theta$ is the direction of the truck’s displacement from the warehouse from south of east.

$\theta=tan^{-1}(\dfrac{2.45}{3.2})$

$\theta=37.43^{\circ}$

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

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3 years ago