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Dimas [21]
3 years ago
10

How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?

Physics
1 answer:
lesya [120]3 years ago
3 0
Force = mass * acceleration = 1500kg * 8m/s²
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A slice of cheese has a mass of 40 g and a volume of 23 cm^3. What is the density of the cheese in units of g/cm^3 and g/mL?
Mila [183]

The mathematical and proportional relationship between mL and cm ^ 3 said us that 1cm ^ 3 is equivalent to 1mL.

If the density is considered as the amount of mass per unit volume we will have to

\rho = \frac{m}{V}

here,

m = mass

V = Volume

Replacing we have that

\rho = \frac{40g}{23cm^3}

\rho = 1.739g/cm^3

As 1mL = 1cm^3 we have that the density in g/mL is,

\rho = 1.739g/mL

6 0
3 years ago
ILL GIVE BRAINLY THING
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Answer:

About 3 trips

Explanation: if we do 2.5m*1.6m*0.75 it equals to 11000 then we divide that to 11m3 and it gives you 3.6 so it will be about 3 times

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The last equation gives you the tension in the string on the right:

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4 0
2 years ago
A circuit breaker is tripped when the
SCORPION-xisa [38]

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Explanation:

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A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
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