1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
miss Akunina [59]
3 years ago
14

3. A snowmobile has a mass of 2.25 x 102 kg. A constant force is exerted on it for 50.0 s. The snowmobile’s initial velocity is

7.00 m/s and its final velocity is 30.0 m/s.
a. What is its change in momentum?
b. What is the magnitude of the force exerted on it?
Physics
1 answer:
Art [367]3 years ago
3 0

Answer:

a. m(v-u)

momentum =5175kgm/s

b. 103.5N

You might be interested in
What clues are useful in reconstructing pangaea
Gnoma [55]
You can use map and notice one thinh. If you flipp over the edges of continents and put them together, you will get a big single continent that is called pangaea. Practically it's impossible but it could be imagined.
6 0
4 years ago
Read 2 more answers
Which object would have the most momentum?
soldier1979 [14.2K]

Answer:

B

Explanation:

Momentum is the product or multiplication of a body's mass and its speed.

Since all options have the mass and speed in the same units, there is no need for conversion.

A. 20 x 500 = 10000,  B. 200 x 60 = 12000

The same goes for the rest!

6 0
3 years ago
Read 2 more answers
I need the answer to number #23 please will give BRAINLIEST
Murrr4er [49]
Yeah lowkey i think it’s B
3 0
3 years ago
I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove
Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
     1/2 ( 0.466697 + 0.307499) = 0.387 098  AU
 
This also happens to be 1/2 of the major axis of the elliptical orbit.


3 0
4 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
Other questions:
  • An athlete always runs before taking a jump . why?
    6·1 answer
  • A human eye is an example of a simple machine found in the human body. True or false
    11·1 answer
  • Why electric potential of earth is taken to be zero?
    15·1 answer
  • The Starship Enterprise speeds past an asteroid at 0.920c. If an observer on the asteroid sees 10.0 seconds pass on her watch, h
    9·1 answer
  • The weight of an apple near the surface of the Earth is 1 N. What is the weight of the Earth in the gravitational field of the a
    11·1 answer
  • A parallel-plate capacitor is charged until it carries charge + q +q on one plate and charge − q −q on the other plate. The capa
    10·1 answer
  • In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system.
    10·1 answer
  • A gas mixture has 10% O2, 50% Ar (40 gmw) and 40% Pu (244 gmw). What is the density of this mixture?
    8·1 answer
  • Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?​
    10·1 answer
  • The electricity received at an electrical substation has a potential difference of 20,000 V. What should the ratio of the turns
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!