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weqwewe [10]
3 years ago
7

Please help me fast and i brainliest!! Two balls are released from the same height. Ball A is released on the surface of Earth,

and ball B is released on the surface of the moon. The mass and net force on each ball are recorded in the table. Net Force (N) Mass (kg) Ball A 19.6 2 Ball B 9.6 6 Ball C 6.6 ? Which statement is correct about the mass of ball C, which is released on the surface of the moon from the same height? The mass of ball C is less than the masses of ball A and ball B. The mass of ball C cannot be determined from the information provided. The mass of ball C is greater than the masses of ball A and ball B. The mass of ball C is greater than the mass of ball A but less than the mass of ball B
Physics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

If they both start at the same height then most likely that means that each has an intical velocity  and a net force sooooo ball A 19.6

Explanation:

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If you are driving 95 km????h along a straight road and you look to the side for 2.0 s, how far do you travel during this inatte
Gala2k [10]

Answer:

52.7 m

Explanation:

Given that

speed of the vehicle, v = 95 km/h

time of inattentiveness, t = 2 s

distance travelled, s = ?

Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.

95 km/h = 95 * 1000/3600 m/s

95 km/h = 95000/3600 m/s

95 km/h = 26.38 m/s

Now, we use our derived speed in m/s

Speed of a moving vehicle is given by,

v = s/t, where

v = speed in m/s

s = distance travelled, in m

t = time spent, in s

if we make d the subject of formula by rearranging the equation, we have

s = v * t

distance travelled, s = 26.38 * 2

distance travelled, s = 52.7 m

therefore, during this inattentive period, 52.7 m was travelled.

3 0
3 years ago
List and describe the top 3 qualities of an<br> effective leader?
s2008m [1.1K]

BEING HONEST , OPTIMISTIC,DISCIPLINED, PUNCTUAL ,TRUTHFUL AND GOOD ORATOR(SPEAKER) , AND HOPEFUL ARE OTHER BASIC QUALITIES AN EFFECTIVE LEADER MUST POSSESS. ALSO,HE MUST NEVER DIFFERENCE BETWEEN ANY OF HIS FOLLOWERS OR DO PARTIALLY AT ANY TIME . HE SHOULD TREAT EVERY ONE EQUALLY.

Explanation:

7 0
4 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
3 years ago
In what way are mercury and venus similar ?
Neporo4naja [7]

Answer:

they are both planets they are both made of rock and  metel

Explanation:

3 0
4 years ago
Read 2 more answers
To find the acceleration of a glider moving down a sloping air track, you measure its velocities (V1 and V2) at two points and t
jeka94

Answer:

(a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

Explanation:

Given that,

Initial velocity v_{1}=0.21\pm 0.05\ m/s

Final velocity v_{2}=0.85\pm 0.05\ m/s

Time t = 8.0\pm 0.1\ sec

(a). We need to calculate the average acceleration

Using formula of acceleration

a=\dfrac{v_{2}-v_{1}}{t}

Put the value into the formula

a=\dfrac{0.85-0.21}{8.0}

a=0.08\ m/s^2

(b). We need to calculate the uncertainty in the velocity

Using formula of the uncertainty

\dfrac{\Delta v}{v}=\dfrac{\Delta v}{v_{2}-v_{1}}

Put the value into formula

\dfrac{\Delta v}{v}=\dfrac{0.05+0.05}{0.85-0.21}

\dfrac{\Delta v}{v}=0.15625

We need to calculate the uncertainty in the time

Using formula for time

\dfrac{\Delta t}{t}=\dfrac{0.1}{8.0}

\dfrac{\Delta t}{t}=0.0125

We need to calculate the uncertainty in the acceleration

Using formula for acceleration

\dfrac{\Delta a}{a}=\dfrac{\Delta v}{v}+]\dfrac{\Delta t}{t}

Put the value into the formula

\dfrac{\Delta a}{a}=0.15625+0.0125

\dfrac{\Delta a}{a}=0.16875

\Delta a=0.16875\times0.08

\Delta a=\pm0.0135

Hence, (a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

3 0
3 years ago
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