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
dybincka [34]
3 years ago
15

35. A daredevil is launched out of a

Physics
1 answer:
Nezavi [6.7K]3 years ago
6 0

Answer:

845 m

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 13 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (13 s) + ½ (10 m/s²) (13 s)²

Δy = 845 m

You might be interested in
If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa
vovikov84 [41]
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
  
          F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation. 
7 0
3 years ago
In an isolated system, Bicycle 1 and Bicycle 2, each with a mass of 10 kg,
dimulka [17.4K]

Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

momentum before the crash will be:

M1V1 - M2V2

Substitute all the parameters into the formula

10 × 5 - 10 × 3

50 - 30

20 kgm/s

Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s

7 0
3 years ago
Why do you need to learn about Volcanoes?
nevsk [136]
That they sometimes explode?
5 0
3 years ago
Read 2 more answers
To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

W = 11.5* \frac{\pi}{3}

W = 12.09J

4 0
3 years ago
I dont know how to do this at all please help
worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
Earth's surface.

In other words, if I  were standing on the surface of that neutron star,
I would weigh  1.82 x 10¹² tons, give or take.     
3 0
3 years ago
Other questions:
  • An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular
    9·1 answer
  • La frecuencia de una onda es 60Hz y su velocidad, 30 m/s. Calcula su longitud de onda
    9·1 answer
  • A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans
    14·2 answers
  • Discuss potential behavioral concerns for people should they travel to Mars
    6·1 answer
  • The net force on an object is 200 N. The object accelerates at 10 m/s? What is the mass of the object ?
    5·1 answer
  • A force of 50 N was necessary to lift a rock. A total of 150 J of work was done. How far was the rock lifted? ​
    15·1 answer
  • The chemical formula for Glucose is C12H6012.<br> True<br> False
    8·2 answers
  • Which of the following will cause water to change to ice!
    6·2 answers
  • What is the pulling force of this train, if its engine provides 4200 kW power and 90 km/h speed?
    11·1 answer
  • What is the coordination number of each atom in Silicon?
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!