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2 years ago

Uncle Harry weighs 80 newtons. What is his mass in kilograms?

Physics

2 answers:

vivado [14]2 years ago

8 0

Answer:

8.1577297038

Explanation:

pls give brainlest

Readme [11.4K]2 years ago

5 0

Answer:

8.158 Kilograms Force

Explanation:

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A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

bonufazy [111]

Answer:

(a) $P=33000W$

(b) $P=51000W$

Explanation:

The average power is defined as the amount of work done during a time interval:

$P=\frac{W}{t}(1)$

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

$W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)$

Recall that the weight is given by:

$w=mg\\m=\frac{w}{g}(3)$

The car accelerates uniformly from rest ( $v_0=0$ ). Replacing (3) in (2), we have:

$W=\frac{wv_f^2}{2g}$

(a) Finally, we replace this in (1):

$P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W$

(b)

$P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W$

3 0

3 years ago

A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency

Vikki [24]

Answer:

Explanation:

f = $\sqrt{T/(m/L)} / 2L$

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

(wow that's massive for a "rope")

f = $\sqrt{120/12} /(2(3))$ )

f = $\sqrt{10\\}$ /6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

5 0

2 years ago

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

3 years ago

A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,

Alika [10]

Explanation:

Given that,

Displacement in ice block, $d=14i-11j$

Force exerted by water, $F=158i-179j$

To find,

Work done by the force during the displacement.

Solve,

We know that the product of force and displacement is called work done. It is also equal to the dot product of force and displacement as :

$W=F.d$

$W=(158i-179j).(14i-11j)$

We know that, i.i = j.j = k.k = 1

$W=2212+1969=4181\ J$

So, the work done by the force on the block during the displacement is 4181 Joules.

7 0

3 years ago

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

3 years ago