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3 years ago

A tiger runs at 58 km/h [S]. What is the displacement of the tiger in 38 s?

Physics

1 answer:

Artyom0805 [142]3 years ago

3 0

58 K/h = 58000/3600= 16.1 m/s
In 38 s displacement is 38x16.1= 612.2 m

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Ket [755]

Diagram 4 is the correct answer.

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3 years ago

A suitcase (mass m = 18 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures

Sindrei [870]

Answer:

P=2736 Pa

Explanation:

According to Newton we have that:

∑ $F=m*a\\$

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

∑ $F=m*g+m*a=m*(g+a)$

$F=205.2N$

We know that the pressure is given by:

$P=\frac{F}{A}\\P=\frac{205.2N}{(0.50m)*(0.15m)}\\P=2736N/m^2=2736Pa$

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3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

irinina [24]

288.51 N is the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The horizontal component of force + the vertical component of force

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.

Learn more about components of forces here brainly.com/question/26446720

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2 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Orlov [11]

Answer:

I dont know this but I have an app for you if your really strangling

Explanation:

its called Qanda. it really good and teachers help you a lot

8 0

3 years ago

To pull a 57 kg crate across a horizontal frictionless floor, a worker applies a force of 230 N, directed 34° above the horizont

lisabon 2012 [21]

Answer:

$a. W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d. W_t=235\ J$

Explanation:

Given: that,

Angle of inclination of the surface, $\theta=34^{\circ}$

mass of the crate, $m=57\ kg$

Force applied along the surface, $F=230\ N$

distance the crate moves after the application of force, $s=1.1\ m$

a) work done = F× s

work done = 230 × 1.1

work done = 253 J

b) Work done by the gravitational force:

$W_g=m.g\times h$

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

$h=s\times sin\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m$

So, the work done by the gravity:

$W_g=57\times 9.8\times (-0.615) \\= - 343.54 J$

∵direction of force and displacement are opposite.

= - 343.54J

The normal reaction force on the crate by the inclined surface:

$F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N$

Total work done on crate is with respect to the worker:

$W_t=235\ J$

5 0

3 years ago