Ask question

Ask question

All categories

1 year ago

5

A woman takes her dog rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

rizontal sidewalk. What force is tending to lift Rover vertically off the ground?
Please help I'm stuck.

1 answer:

IrinaK [193]1 year ago

4 0

19.8 N force is tending to lift Rover vertically off the ground.

<h3>What is horizontal and vertical component?</h3>

The horizontal velocity component ( $v_{x}$ ) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component ( $v_{y}$ ) describes the influence of the velocity in displacing the projectile vertically.

According to the question,

The women pulls the dog with a force of 30 N at an angle of 29° from the horizontal.

Horizontal component= 30cos(29°) = 22.2 N

Vertical component = 30sin(29°) = 19.8 N

Therefore,

The horizontal component would tend to make the dog move forward and the vertical component would tend lift it off the ground.

Hence,

19.8 N force is tending to lift Rover vertically off the ground.

Learn more about horizontal and vertical component here:

brainly.com/question/11776718

#SPJ1

You might be interested in

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

Ugo [173]

Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

So

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

So

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

6 0

3 years ago

What is the equivalet tempreture of 0 kelvin in celcius scale <br> HELP NEED IT IN 5 MIN

Alona [7]

Answer:

-273.15 degree celcius

Explanation:

7 0

3 years ago

Gravitational energy can be negative?

AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

6 0

3 years ago

Which statements describe nuclear reactions? Check all that apply.

Alexandra [31]

Explanation:

Nuclear reactions are the reactions in which nucleus of an atom changes either by splitting or joining with the nucleus of another atom.

There are two types of nuclear reactions.

Nuclear fission - In this process, large atomic nuclei splits into smaller nuclei.
Nuclear fusion - In this process, two small nuclei combine together to form a large nuclei.

Both nuclear fission and fusion processes involve nuclei of atoms.

For example, $^{0}n + ^{235}_{92}U \rightarrow ^{236}_{92}U \rightarrow ^{144}_{56}Ba + ^{89}_{36}Kr + 3n + 177 MeV$

Thus, we can conclude that statements which are true are as follows.

Nuclear reactions involve the nuclei of atoms.

The products of nuclear reactions are lighter than the reactants.

7 0

3 years ago

Read 2 more answers

Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit

riadik2000 [5.3K]

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

$h_1 = 138.99 Btu/lbm$

steady flow energy equation is givena s

$h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}$

$138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}$

$h_2 = 94.344 Btu/lbm$

At 90 lb/in2 Tsat = 72.78 degree F

$h_f = 35.715 Btu/lbm$

hfg = 77.345 Btu/lbm

h = hf + x hfg

$94.344 = 35.715+ x \times 77.345$

solving for x we get

x = 0.75801 = 75.801%

$T_2 = 72..78 degree F$

5 0

3 years ago