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djyliett [7]
3 years ago
5

a car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. when the brakes are applied, it takes the ca

r 3.0 seconds to stop. what is the force required to stop the car?
Physics
1 answer:
Evgen [1.6K]3 years ago
5 0

It i believe it would be 7.3 × 10∧3.

correct me if im wrong

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Calculate the force exerted by a mental ball having a mass of 70kg moving with speed of 20m/s>2
lord [1]

Answer:

F = 1400 N

Explanation:

It is given that,

Mass of the ball, m = 70 kg

It is moving with an acceleration of 20 m/s². We need to find the force exerted by the ball.

Force is given by the product of mass and acceleration. So,

F = ma

F=70\ kg\times \ 20m/s^2\\\\F=1400\ N

So, the force of 1400 N is exerted by a metal ball.

8 0
3 years ago
Observe the given figure and find the the gravitational force between m1 and m2.​
Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0
3 years ago
Protons are the smallest particle of an element that retains the original Characteristics of the element. True or False
dlinn [17]

Answer:

False; that’s an atom

Explanation:

5 0
3 years ago
Read 2 more answers
Which instrument would make rice vibrate easier, a tuba or a flute? Explain why. Hint: think about the difference between high a
algol13

Answer:

I assume the higher notes would make the rice vibrate more easily, so a flute.

8 0
3 years ago
Help me please, need more assistance
Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0
3 years ago
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