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zubka84 [21]
3 years ago
8

Which of the following statements are not true about gravity? Check all that apply.

Physics
2 answers:
vladimir1956 [14]3 years ago
7 0

D and c, I believe are the answers to this.

liberstina [14]3 years ago
3 0

C gravity only exists only on earth

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5) Sally has a mass of 50 kg. She ran with a velocity of 20 m/s.
Ket [755]

Answer:

<h2>10,000 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2} \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 50 \times  {20}^{2}  \\  = 25 \times 400 \\  = 10000

We have the final answer as

<h3>10,000 J</h3>

Hope this helps you

7 0
2 years ago
A snowball accelerates at
ANTONII [103]

Answer:

0. 1226495726kg

Explanation:

Force is the product of mass and acceleration.

Mathematically,

Force(F) = mass (m)×acceleration(a)

Substituting the values into the equation

2. 87=m×23. 4

2. 87=m (23. 4)

2. 87/23. 4=m (23. 4)/23. 4

2. 87/23. 4=m

0. 1226495726=m

8 0
3 years ago
Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and
Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

v_f^2=v_i^2+2ax

Where,

v_f = Final velocity

v_i = Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

F_f = \mu_k (mg)  \rightarrowFrictional Force

F = ma \rightarrow Force by Newton's second Law

Where,

m = mass

a= acceleration

\mu_k = Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

F_f = F

\mu_k mg=ma

a = \mu_k g

Therefore,

v_f^2=v_i^2+2ax

0=v_i^2+2(\mu_k g)x

Re-arrange to find x,

x = \frac{v_i^2}{2(-\mu_k g)}

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0
3 years ago
In which situation is maximum work considered to be done by a force?
umka2103 [35]

Answer:

A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.

W = F•x cos ∅

If ∅ = 0°,

W = F•x ===> Maximum Work Done.

If ∅ = 45°,

W = F•x/√2

If ∅ = 90°,

W = 0

If ∅ = 180°,

W = –F•x ===> Minimum Work Done.

7 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0
2 years ago
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