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
lesya [120]
3 years ago
11

What is the momentum of a 8 kilogram cart moving with a speed of 12 meters per second?

Physics
1 answer:
lisov135 [29]3 years ago
5 0

Answer:

Explanation:

momentum = mv

m = 8 kg

v = 12 m/s

momentum = 8 * 12 = 96 kg m/s

You might be interested in
Terry usually rides his bike at 15mph. If his speed is reduced by 3 mph , how far can he ride in 1.7 hours ?
kakasveta [241]
105 miles because you have to use the gif arable
7 0
3 years ago
28 points
dem82 [27]
Answer: C. 1.64 x 10-3 m/s2
7 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0
3 years ago
. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
Two positive point charges repel each other with force 0.36 N when their separation is 1.5 m. What force do they exert on each o
egoroff_w [7]
The answer is: 0.81

I hope this helps :)
6 0
3 years ago
Other questions:
  • What are some predicted affects of climate change linked to global warming
    7·1 answer
  • What will change the velocity of a periodic wave?
    10·1 answer
  • A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h
    8·1 answer
  • A negative externality:
    6·1 answer
  • A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i
    11·1 answer
  • A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a
    9·1 answer
  • A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the
    9·1 answer
  • Two particles, with identical positive charges and separation of 2.55 10-2 m, are released from rest. Immediately after the rele
    11·1 answer
  • Long-Distance Space Travel
    7·2 answers
  • ___________________ is the opposition to the flow of electric charge. (Resistance or Electric power)
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!