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4 years ago

You perform 556 J of work lifting a box to a height of 1.3m. How much force did you use to lift the box?

Physics

2 answers:

yuradex [85]4 years ago

5 0

Answer:

The work is given 556 J. Here work is performed horizontally so your distance d=1.3 m

W=F.d

F=W/d =556/1.3 = 427.69 N

Explanation:

babunello [35]4 years ago

4 0

The work is equal to the product between the force applied and the distance covered by the box:
$W=Fd$
In our problem, W=556 J, and d=1.3 m (the box is lifted to a height of 1.3 m, so it covered 1.3 m from its initial point). Therefore we can find the force applied to lift the box:
$F= \frac{W}{d}= \frac{556 J}{1.3 m}=427.7 N$

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viktelen [127]

Given:

Momentum of the dog (p) = 120.5 kg m/s

Speed of the dog (v) = 5 m/s

To Find:

Mass of the dog (m)

Concept/Theory:

$\underline{\underline{ \bf{\Large{Momentum}}}}$

It is defined as the quantity of motion contained in a body.
It is measured as the product of mass of the body and it's speed.
It is represented by p.
It's SI unit is kg m/s
Mathematical Representation/Equation of Momentum: $\boxed{ \bf{p = mv}}$

Answer:

By using equation of momentum, we get:

$\rm \longrightarrow m = \dfrac{p}{v} \\ \\ \rm \longrightarrow m = \dfrac{120.5}{5} \\ \\ \rm \longrightarrow m = 24.1 \: kg$

$\therefore$ Mass of the dog (m) = 24.1 kg

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3 years ago

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm

mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

$f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}$

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

$V_{sph}=\frac{4\pi}{3}r^3$

So the diameter is:

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3 years ago

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Anon25 [30]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between

mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

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V = 0.4061 V

The potential difference between the plates is 0.4061 V

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3 years ago

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

$f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz$

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3 years ago