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Finger [1]
2 years ago
12

Select the correct answer from the drop-down menu. anne applies a force on a toy car and makes it move forward. what can be said

about the forces acting on the toy car at the moment anne applies the force? the forces acting on the toy car are .
Physics
1 answer:
nalin [4]2 years ago
6 0

Answer:

The forces could be gravity, friction between the car and the ground, the force Katie is applying and the normal reaction.

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How are systems different from industries? Use an example to support your answer.
jeka94

Answer:

Explanation:

An industrial system consists of inputs, processes and outputs. The inputs are the raw materials, labor and costs of land,transport, power and other infrastructure. The processes include a wide range of activities that convert the raw material into finished products.

7 0
2 years ago
Read 2 more answers
A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6
Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is  -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ =  4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} =  \frac{-461}{92} m/s

v₃ = -5.011 m/s.

3 0
3 years ago
Read 2 more answers
Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont
SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

T = \frac{2 u Sin\theta }{g}

T = \frac{2 \times 20 \times Sin35 }{9.8}

T = 2.34 second

(b) The formula for the maximum height is given by

H = \frac{u^{2} \times Sin^{2}\theta }{2g}

H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}

H  = 6.71 m

(c) The formula for the range is given by

R = \frac{u^{2} \times Sin 2\theta }{g}

R = \frac{20^{2} \times Sin 2 \times 35}{9.8}

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0
3 years ago
Find the acceleration of the blocks when the system is released. The coefficient of kinetic friction is 0.4, and the mass of eac
grin007 [14]

Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

Really not possible without a complete setup.

I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g

                                     F = ma

mg - mgsinθ - μmgcosθ = (m + m)a

      mg(1 - sinθ - μcosθ) = 2ma

      ½g(1 - sinθ - μcosθ) = a

maximum acceleration is about 2.94 m/s² when θ = 0

acceleration will be zero when θ is greater than about 46.4°

8 0
3 years ago
Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two
likoan [24]

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

6 0
3 years ago
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