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xeze [42]
3 years ago
14

Which of the following is an example of matter?

Physics
2 answers:
krek1111 [17]3 years ago
8 0

Answer:

(D) all of the above

Explanation:

All of the choices describe matter.

Gemiola [76]3 years ago
5 0
D cause I learned this in the 5th grade and some how remember this (I think because it’s all up in space)
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What is the average acceleration of a jet ski that goes from rest to 10 m/s west in 4 s? Take the direction west to be positive,
skad [1K]

|acceleration| = (change in speed) / (time for the change)

                       = (10 m/s - 0)          /  (4 s)

                       =        (10 / 4) (m/s²)

                       =            2.5 m/s² .

The direction of the acceleration is west.
3 0
3 years ago
At what temperature is the average kinetic energy of an atom in helium gas equal to 6.21 x 10-21 j?
NARA [144]

The average kinetic energy of an atom in helium gas at temperature T is given by,

K.E=\frac{3}{2}KT

here, K is botzmann constant =1.38×10⁻²³ m² kg s⁻² K⁻¹

Given, K.E= 6.21×10⁻²¹ J

Substituting the values,

6.21*10^{-21}=\frac{3}{2}1.38*10^{-23}T

T= 300 K

Therefore, at temperature 300 K the average kinetic energy of the atom in helium gas is 6.21×10⁻²¹ J.

5 0
4 years ago
HELP ME IM IN 3TH GRADE AND ONLINE SCHOOL IS POOP
snow_lady [41]

Answer:

1. evaporation

2. sublimation

3. melting

4. freezing

5. deposition

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7 0
3 years ago
PLEASEEEE HEEEEELP!!!!!
8090 [49]

Answer:

Before:

p_{truck}=16400\ kg.m/s

p_{car}=10000\ kg.m/s

After:

p_{truck}=8000\ kg.m/s

p_{car}=8400\ kg.m/s

v_{fcar}=8.4\ m/s

F=9333.33 \ Nw

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

p_1=m_1v_{1o}+m_2v_{2o}

After the collision, they have speeds of v1f and v2f and the total momentum is

p_2=m_1v_{1f}+m_2v_{2f}

Impulse J is defined as

J=F.t

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

J=\Delta p

As the total momentum is conserved:

p_1=p_2

m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}

We can compute the speed of the second object by solving the above equation for v2f

\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{  m_2 }

The given data is

m_1=2000\ kg

m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s

a) The impulse will be computed at the very end of the answer

b) Before the collision

p_{truck}=2000\cdot 8.2=16400\ kg.m/s

p_{car}=1000\cdot 0=0\ kg.m/s

c) After collision

p_{truck}=2000\cdot 4=8000\ kg.m/s

Compute the car's speed:

\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }

v_{2f}=8.4\ m/s

And the car's momentum is

p_{car}=1000\cdot 8.4=8400\ kg.m/s

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s

The force can be computed as

\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw

The force on the car has the same magnitude and opposite sign

7 0
4 years ago
Please help! Question is attached.​
il63 [147K]

Answer:

a. 6

b. 6 m/s²

c. 300 m to the right

d. 30 secs

Explanation:

slope = rise /run

60-0/10-0

= 6

b. slope = acceleration = 6 m/s²

c. d=ut+1/2at²

t=10 (segment A last for 10 secs)

u - initial velocity = 0

so d = 0(10)+1/2*6*10²

=300 m

6 0
3 years ago
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