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
nignag [31]
3 years ago
7

What is the acceleration of softball if it has a mass of 5 kg and hits the catchers glove with a force of 35 N

Physics
2 answers:
nekit [7.7K]3 years ago
4 0

Answer:

7 m/s²

Explanation: Force = Mass x acceleration  

Also note that 1 N = 1 kg.m/s²

a=F/ m

a= 35 N/ 5 kg = 35 kg.m/ s²  ÷  kg                 Aside:( so you cross out the kg)

a= 7 m/s²

Katen [24]3 years ago
3 0

Answer:

50m/s2^

Explanation:

We know , by newton's law of motion .

F=ma

a=f/m

a=25/0.5

a=50m/s2^

Therefore , the acceleration of a softball is  .

You might be interested in
A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a
myrzilka [38]

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

7 0
3 years ago
I need help ASAP please :)
rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0
2 years ago
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting
Eddi Din [679]

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2

where, g is the acceleration due to the gravity

on substituting the values, we get

\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

7 0
3 years ago
What might you have if you dont feel well?
snow_lady [41]
You may have a cold if you do not feel well, depends on the symptoms
3 0
3 years ago
Read 2 more answers
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
Other questions:
  • The force exerted by electrically changed objects
    8·1 answer
  • A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou
    8·1 answer
  • How to find current in a series circuit
    10·1 answer
  • What happens when the temperature of the air cools?
    13·1 answer
  • A small lightbulb is 1.06 m from a screen. A) If you have a convex lens with 20 cm focal length, where are the two lens location
    12·1 answer
  • Which is an example of projectile motion?
    5·2 answers
  • In Newtons famous event that why, apple always falls to the Earth. Suppose the weight of the apple is 2.5 N. Then calculate the
    11·1 answer
  • 22. White noise is ...
    8·1 answer
  • All living things inherit similar traits from their
    7·1 answer
  • ILL GIVE BRAILIST THING
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!