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mafiozo [28]
3 years ago
5

A golf ball is at rest on the grass when a golfer walks up and applies 10 N of force to it. There is 2 N of friction, resulting

in 8 N of net force. What will happen to the ball? (4 points)
It will remain at rest.

It will move in the direction opposite the net force.

It will move slowly at first, then speed up.

It will move in the direction of the net force.
Physics
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

The forth one

Explanation:

8N of net force is applied to the golf ball

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uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th
adoni [48]

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

7 0
3 years ago
A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?
MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

8 0
3 years ago
A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0
3 years ago
Read 2 more answers
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
During which type of collision is none of the energy converted to heat?
Galina-37 [17]

Answer:

D.

Explanation:

7 0
4 years ago
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