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Lelu [443]
4 years ago
12

A lizard ran 3 meters from his rock to his friend's house. He ran

Physics
2 answers:
lbvjy [14]4 years ago
5 0

Answer:

4.5 meters

Explanation:

3 meters to his friend's house. +3

Halfway back. +1.5

3+1.5

=4.5

4.5 meters in total

Leya [2.2K]4 years ago
4 0

Answer:

Distance traveled, d = 4.5 meters

Explanation:

It is given that, a lizard ran 3 meters from his rock to his friend's house, distance covered by lizard, d₁ = 3 m

He ran  back halfway and stopped, now the distance covered, d_2=\dfrac{3}{2}=1.5\ m

We need to find the distance traveled. The overall path covered by an object during its entire journey is called distance covered.

To find the distance covered by the lizard we need to simply add both distances as :

d=d_1+d_2

d=3+1.5                      

d = 4.5 meters

So, his distance traveled is 4.5 meters. Hence, this is the required solution.                                          

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maxonik [38]

Answer:

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Explanation:

8 0
3 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
choli [55]

Answer:

a)  F = 1.26 10⁵ N, b)  F = 2.44 10³ N, c)   F_net = 1.82 10³ N  directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

         I = Δp

         F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

         F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

         F = m v₀ / t

         F = 63.5 7.89 / 3.99 10⁻³

         F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

          F = 63.5 7.89 / 0.205

          F = 2.44 10³ N

c) The net force is

         F_net = Sum F

         F_net = F - W

         F_net = F - mg

let's calculate

         F_net = 2.44 10³ - 63.5 9.8

         F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

6 0
3 years ago
Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha
Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y=  F₂ cos 32° j

F₂y=  35 x cos 32° j =  29.68 j

The total force F

F= 33.47 i +   18.54 i - 55.71 j +   29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

F=\sqrt{52.01^2+26.03 ^2}\ N

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0
4 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 7.5 oz . Aluminum has a density of 2.70 g/cm3. W
Tems11 [23]

Answer:

The thickness of the foil is 0.017 mm.

Explanation:

Given that,

Weight = 7.5 oz = 212.6175 gm

Density = 2.70 g/cm³

Area of aluminium = 50 ft² = 46451.52 cm²

We need to calculate the thickness of the foil

Using formula of density

\rho=\dfrac{m}{V}

\rho=\dfrac{m}{A\times t}

t=\dfrac{m}{A\times \rho}

Where, A = area

t = thickness

m = mass

Put the value into the formula

t=\dfrac{212.6175}{46451.52\times2.70}

t=0.00170\ cm

t=0.017\ mm

Hence, The thickness of the foil is 0.017 mm.

3 0
4 years ago
Objects a and b each have a mass of 25 kilograms. object a has a velocity of 5.98 meters/second. object b is stationary. they un
mash [69]

The total kinetic energy of the system after collision is 223.5J

In elastic collision kinetic energy and momentum are conserved.

According to the question

               mass of object a = 25kg

               mass of object b =  25 kg

        initial velocity of a u1 = 5.98 m/s

        initial velocity  of b u2 = 0

so from momentum conservation-

       mau1 + mbu2 = (m1+m2)v

       25kg × 5.98m/s + 25×0 = (25+25)v

              v = 2.99 m/s

Now the total kinetic energy after the collision will be:

          final kinetic energy = 1/2 (m1+m2) v²

                                           = 1/2 (25+25)× (2.99)²

                                           = 223.5 J

    Thus, total kinetic energy of the system after collision is 223.5J

Learn more about elastic collision here:

  brainly.com/question/1808045

    #SPJ4

7 0
2 years ago
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