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aksik [14]
2 years ago
8

Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a

force of 0.035 N. The water balloon has a mass of 11.4 grams. How fast will Sachi have to make the water balloon accelerate to hit the target with enough force and win the prize?
Physics
2 answers:
Verizon [17]2 years ago
8 0
3.1 the only reason i know this is cause i got it wrong 
Mashutka [201]2 years ago
8 0
Remember that acceleration (a) equals force (F) over mass (m) or a=F/m. In this equation, both the force, 0.035N, and the mass, 11.4 grams, are provided. However, the mass is provided in the wrong unit, as it should be calculated in kilograms. To set up this equation, you will have to convert 11.4 grams into .0114 kilograms then divide .035 N over 0.114 kg to get your answer of 3.07 m/s^2 rounded to 3.1 m/s^2
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A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it
lesya692 [45]

Answer:

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

\theta = \omega t

now the position vector at a given time is

r = Rcos\theta \hat i + R sin\theta \hat j

now the linear velocity is given as

v = \frac{dr}{dt}

v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

6 0
1 year ago
While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe
kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

P_{Miy} + p_{Piy} + j_{y}= P_{Mfy} +P_{pfy}

m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sinФ

60 ·8 + 0 = (60 + 80)V_{f}sinФ

480 = 140V_{f} sinФ................. (I)

y-axis component form (+y north);

P_{Mix} + p_{Pix} + j_{x} = P_{Mfx}+ P_{pfx}

m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cosФ

0 + 80.9 = (60 + 80)V_{f}cosФ

 720= 140V_{f}cosФ

140Vf=\frac{720}{cos}Ф......................................(2)

 Substituting (2) into (1) to give the angle;

 480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

 

7 0
2 years ago
Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend
Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

6 0
2 years ago
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valentina_108 [34]

the sentence with leeward side and the sentence that has windward side both have errors.

5 0
2 years ago
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