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geniusboy [140]
3 years ago
14

How much net force is needed to accelerate a 15 kg mass at 2.8 m/s^2

Physics
1 answer:
lord [1]3 years ago
8 0
From the information given, there is a 15 kg mass that needs to be accelerated to a velocity of 2.8m/s² We use the formula: F = m × v. Where F = force, m = mass and v = velocity. In our case, m = 15kg and v = 2.8m/s². Therefore: F = m × v. F = 15 × 2.8. F = 42. Therefore the net force needed to accelerate the 15kg mass to 2.8m/s² is 42 kg m/s² (or 42 Newtons)
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Estimate the radiation pressure due to a bulb that emits 25 W {\rm W} of EM radiation at a distance of 9.5cm \;cm from the cente
madam [21]

Answer:

7.3*10^{-7}  \frac{N}{m^2}

Explanation:

\frac{25}{(4*\pi*(0.095)^{2}*3*10^8 }

Hope this is right and helps!

8 0
3 years ago
A car starts from rest at the top of a hill with 45 J of gravitational
kherson [118]

Answer:

<em>The car will be moving at 5.48 m/s at the bottom of the hill</em>

Explanation:

<u>Principle of Conservation of Mechanical Energy</u>

In the absence of friction, the total mechanical energy is conserved. That means that

E_m=U+K is constant, being U the potential energy and K the kinetic energy

U=mgh

\displaystyle K=\frac{mv^2}{2}

When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:

\displaystyle \frac{mv^2}{2}=45

Multiplying by 2:

\displaystyle mv^2=90

Dividing by m:

\displaystyle v^2=\frac{90}{m}

Taking square roots:

\displaystyle v=\sqrt{\frac{90}{m}}

\displaystyle v=\sqrt{\frac{90}{3}}

v=\sqrt{30}

v = 5.48 m/s

The car will be moving at 5.48 m/s at the bottom of the hill

3 0
3 years ago
In which situation is the maximum possible work done?
mina [271]
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
7 0
3 years ago
A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.0100
natita [175]

Answer:

870N

Explanation:

momentum techniques

7 0
4 years ago
4. An aluminium bar weighs 17 kg in air. How much force is required
kkurt [141]

Explanation:

1ml = 2.7g

Xml = 1.5g

Divide 1.5 by 2.7 to find X.

Obviously, since 1.5 is less than 2.7, you know the answer will be less than 1.

(it’s .5555555555)

4 0
3 years ago
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