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4 years ago

Please help me with my Physical Science! 50 POINTS

Physics

2 answers:

Effectus [21]4 years ago

8 0

Answer:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia.

The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum (mass times velocity) per change in time.

The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A.

Explanation:

I hope this helps:) this is only for answer 1.

DaniilM [7]4 years ago

6 0

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

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A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc

amid [387]

A= f/m

a= 19/2

a= 9.5m/s^2

8 0

3 years ago

A particle in a 799=m-long linear particle accelerator is moving at 0.875C. How long does the particle accelerator appear to the

barxatty [35]

the answer is 386m because m = pi mc=2.23

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2 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago