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natta225 [31]
4 years ago
15

WILL GIVE 5 STARS!!!! HELP ASAP!!!

Physics
1 answer:
pishuonlain [190]4 years ago
3 0
All of the are true facts but because it’s asking how it’s related to the surface layer the only one that mentions about the surface would be A
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What is the mass of an object that has a weight of 80.0 N?
Ira Lisetskai [31]

Explanation:

SUPONIENDO QUE LA ACELERACIÓN DE LA GRAVEDAD ES 9.80 m/s^{2}

USANDO LA SEGUNDA LEY DE NEWTON:

<em>m</em> = 80.0 N/9.80 m/s^{2} = 8.16 kg

4 0
3 years ago
Read 2 more answers
A 2.40 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
klemol [59]

Answer: Hello! Here your answer......

Force = 10.244 Newtons

b) No of oscillations = 0.88

Explanation:

Since the block executes SHM we can write it's position as function of time as

ω is the natural frequency of the system

A is the amplitude of the system

Thus accleration of the block

Thus using the given values at t= 3.50 sec we can calculate the acceleration as

thus force can be calculated using newtons second law as

b)

Now no of oscillations can be obtained as

no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88

Hope this helps! (Brainly)♥

3 0
3 years ago
A ball that is attached to a string travels in a horizontal, circular path, as shown in Figure 1. At time t0 , the ball has a sp
Levart [38]

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

<u>F₁ = 4 F₀</u>

5 0
3 years ago
What two forces act on falling bodies?
monitta
The two forces acting on the object would be 1. Weight because of gravity pulling the object toward the ground and 2. Drag resisting this motion
6 0
3 years ago
A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of
IrinaK [193]

Answer:

A.

Explanation:

Our values are defined by,

m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°

We can express also in a vectorial way, then

v=2\hat{i} \rightarrowno values in \hat{j}

Acceleration as follow,

a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}

We know that velocity is given by,

V(t) = v_0+at \rightarrow v_0 = 2

We need to calculate for t=3, then

v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3

v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})

v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}

v(3) = 7.7\hat{i}+5.7\hat{j}

Our mass is 1000Kg, so the momentum is

P = mv

P= 1000(7.7\hat{i}+5.7\hat{j})

P=7700\hat{i}+5700\hat{j}

7 0
4 years ago
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