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Paha777 [63]
3 years ago
5

Please I really need help on this having a hard time with this please

Physics
1 answer:
swat323 years ago
8 0

There is a force of 2 Newtons to the RIGHT, and a force of 4 Newtons to the LEFT. The Net Force would be 4 Newtons RIGHT minus 2 Newtons LEFT =

B. 2 Newtons to the left

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A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
3 years ago
the freezing point of water is the same as ice's __________ point. a. boiling b. melting c. condensing d. sublimation
nika2105 [10]
Melting. Water freezes at 0 degrees celsius or 32 degrees fahrenheit and therefore water freezes at that temperature or ice melts back to its liquid state at those temperatures.
3 0
4 years ago
As the force applied to an object increases, the acceleration of that object also
Ganezh [65]

Answer:

C. posotive

I dont have any idea to explain my answer.. all I know is that is the aswer

3 0
3 years ago
HELP ASAP!!!!!!! PLEASE WILL GIVE BRAINLIEST
PolarNik [594]

Answer: The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse. If plasma membrane ruptures or breaks down then molecules of some substances will freely move in and out of the cells.

7 0
3 years ago
A particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 3.30 cm. (a) Through what total dista
irinina [24]

Answer:

a) d= 13.2cm

b ) vmax =62.23cm/s

c) amax=1172.81cm/s2

Explanation:

Given:

frequency is 3.00Hz,

the amplitude is 3.30cm.

(a)The total distance travelled by the particle during one cycle is.

d = 4A

d= 4×3.30cm

d = 13.2cm

b) The maximum speed of the particle is calculated as using the formula;

vmax = ωA

= 2πfA

= 2π (3.30cm/s)*(3.0cm/s)

=2× 3.147 × 3.30 × 3.0

vmax =62.23cm/s

The maximum speed occur at mean position(x,0) which means it has maximum speed when it passes through the equilibrium position.

(c) The maximum acceleration of the particle is

calculated below;

amax = ω2A

= (2πf)2A

= (2× 3.142×3)^2×3.30

amax=1172.81cm/s2

The particle has maximum acceleration at the turning points.

5 0
4 years ago
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