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Alexeev081 [22]
3 years ago
15

Where do tadpoles live? a. on land c. both on land and water b. in water d. none of the above

Physics
2 answers:
Bezzdna [24]3 years ago
5 0
Tadpoles live in water
ivanzaharov [21]3 years ago
3 0
Hey :)

Where Do tadpoles live?
Answer:Tadpoles Live in water
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If Vector A is (3, 0) and Vector B is (-3, 3), what is the resultant?
Archy [21]

Answer:resultant vector R = (0, 3)

Explanation: vector A = (3, 0)

vector B =(-3, 3)

Vectors are added such that those in same directions are added together. The resultant vector R is the given by R = (3-3, 0+3)

= (0, 3)

6 0
3 years ago
How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)
leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0
3 years ago
A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h
g100num [7]

Answer:

54 km/hr

Explanation:

m/s to km/hr => 18/5

15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

8 0
3 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi
Elis [28]

Answer: 115.52\ N

Explanation:

Given

Length of plank is 1.6 m

Force F_1 is applied on the left side of plank

Force F_2 is applied 43 cm from the left end O.

Mass of the plank is m=13.7\ kg

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N

Net vertical force must be zero on the plank

\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N

8 0
2 years ago
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