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2 years ago

6

A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

de units in the answer to a numerical question.)

1 answer:

inysia [295]2 years ago

4 0

For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
Rewriting the acceleration as the increment of velocity $\Delta v$ in a time $\Delta t$ : $a= \frac{\Delta v}{\Delta t}$ , F becomes
$F=m \frac{\Delta v}{\Delta t}$
But given the definition of momentum: $p=mv$ , then $m \Delta v$ represents the momentum change. So we can rewrite F as
$F= \frac{\Delta p}{\Delta t}$
And re-arranging the formula we can calculate the value of the change in momentum:
$\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s$

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liberstina [14]

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

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Time taken = 3 seconds

$v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s$

So jumping velocity of Julia = 8 m/s

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2 years ago

Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.

RoseWind [281]

Answer:

A) x _electron = 0.66 10² m , B) x _Eart = 1.13 10² m , C) d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

$R_{E}$ = 6.37 10⁶ m

tex]R_{Sum}[/tex] = 6.96 10⁸ m

Distance = 1.5 10¹¹ m

x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5

x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶

d_sphere = 1.37 10⁻² mm

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2 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0Â° east of

koban [17]

<u>Answer:</u>

Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

Speed of truck = 25 m/s north = 25 j m/s

Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

= (1.43 i + 1.00 j) m/s

Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

Magnitude of velocity = 26.04 m/s

Angle from positive horizontal axis = 86.85⁰

So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

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2 years ago

John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. He is pulling horizontally wi

bulgar [2K]

Answer:

The mass of Laura and the sled combined is 887.5 kg

Explanation:

The total force due to weight of Laura and friction on the sled can be calculated as follows;

$F_T = F_L+F_S$

= (400 + 310) N

= 710 N

From Newton's second law of motion, "the rate of change of momentum is directly proportional to the applied force.

$F_T = \frac{(M_L+M_S)V}{t}$

where;

$M_L$ is mass of Laura and

$M_S$ is mass of sled

Mass of Laura and the sled combined is calculated as follows;

$(M_L+M_S) = \frac{F_T*t}{V}$

given

V = Δv = 4-0 = 4m/s

t = 5 s

$(M_L+M_S) = \frac{710*5}{4}\\\\(M_L+M_S) = 887.5 kg$

Therefore, the mass of Laura and the sled combined is 887.5 kg

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2 years ago

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

Stella [2.4K]

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$ ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

4 0

2 years ago