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True [87]
1 year ago
12

What causes an object to slowdown or speed up

Physics
1 answer:
ivanzaharov [21]1 year ago
7 0

Answer:

A force can speed up or slow down an object. A force can change the direction in which an object is moving. A bigger force on an object will produce a bigger change in the motion. A heavier object requires a larger force than a lighter object in order to undergo the same change in motion.

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Beasy is driving in his car to take care of some errands. The first errand has him
lutik1710 [3]

At this point in the story, Beasy has driven his car (2+6+4) = 12 km.

He is parked at the thrift store, (2+4) = 6 km East and 6 km North of his starting point.

As the crow flies, the thrift store is √(6km² + 6km²) in a straight line from the starting point.

That's √(72 km²) , which works out to 8.485 km .  When rounded to the nearest whole km, he can phone up his wife and tell her he's "eight kilometers from home can you hear me now ?".

Displacement is a vector, so to answer the question completely, we also need to state its direction.

The angle from home to the thrift store, relative to East, is arctan(6km/6km).

That's 45 degrees.  

The full displacement vector is <em>8.485 km Northeast.</em>

3 0
3 years ago
What gravitational force does the moon produce
saul85 [17]

Answer:

1.94\cdot 10^{20} N

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

The gravitational force is always attractive.

In this problem, we have:

m_1 = 5.98\cdot 10^{24}kg is the mass of the Earth

m_2 = 7.34\cdot 10^{22} kg is the mass of the Moon

r=3.88\cdot 10^8 m is the separation between the Earth and the Moon

Therefore, the gravitational force between them is

F=(6.67\cdot 10^{-11})\frac{(5.98\cdot 10^{24})(7.34\cdot 10^{22})}{(3.88\cdot 10^8)^2}=1.94\cdot 10^{20} N

6 0
3 years ago
If a skydiver jumps out of a plane horizontally (in other words with no initial vertical velocity), then what will her vertical
erastovalidia [21]

The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

<h3>Time of motion of the girl</h3>

The time of motion of the girl is calculated as follows;

h = vt + ¹/₂gt²

where;

  • v is initial vertical velocity = 0
  • t is time of motion
  • g is acceleration due to gravity

Substitute the given parameters and solve for time of motion;

50.8 = 0 + ¹/₂(9.8)t²

2(50.8) = 9.8t²

101.6 = 9.8t²

t² = 101.6/9.8

t² = 10.367

t = √10.367

t = 3.22 seconds

<h3>Final vertical velocity of the skydiver</h3>

vf = vi + gt

where;

vi is the initial vertical velocity = 0

vf = 0 + 9.8(3.22)

vf = 31.56 m/s

Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

Learn more about vertical velocity here: brainly.com/question/24949996

#SPJ1

4 0
2 years ago
What is the only component of scalar quantities?
Bad White [126]

Answer:

Scalar quantities have a size or magnitude only and need no other information to specify them. Thus, 10 cm, 50 sec, 7 litres and 3 kg are all examples of scalar quantities.

Explanation:

7 0
3 years ago
Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

4 0
3 years ago
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