Ask question

Ask question

All categories

AleksandrR [38]

2 years ago

6

The human head weighs 4.8 kg and its center of mas 1.8 cm in front of the spinal column joint. If the trapezius muscle inserts 1

.1 cm behind the spinal cord joint, how much downward force does it need to exert in order to maintain equilibrium

2 answers:

lord [1]2 years ago

4 0

Answer:

The downward force does it need to exert in order to maintain equilibrium is 76.974 N

Explanation:

The downward force does it need to exert in order to maintain equilibrium is:

$F=\frac{mgL}{x}$

Where

m = mass of human head = 4.8 kg

g = gravity = 9.8 m/s²

L = center of mass = 1.8 cm = 0.018 m

x = distance = 1.1 cm = 0.011 m

Replacing:

$F=\frac{4.8*9.8*0.018}{0.011} =76.974N$

Anarel [89]2 years ago

3 0

Answer:

77.05N of force

Explanation:

Detailed explanation and calculation is shown in the image below

You might be interested in

A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer

matrenka [14]

Answer:

A) F = - 8.5 10² N, B) I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

v₀ = 8.50 m / s

v_f = -8.50 m / s

F t = m v_f -m v₀

F = $m \frac{(v_f - v_o)}{t}$

let's calculate

F = $1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}$

F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

v = $\sqrt{2g y_o}$

calculate

v = $\sqrt{2 \ 9.8 \ 10}$

v = 14 m / s

to calculate the momentum we use

I = Δp

I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

I = 1.50 (0-14)

I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

I = 21 N s

4 0

1 year ago

The space shuttle is descending through the earth's atmosphere. How is the force of gravity affected?

polet [3.4K]

The answer would be C. It will decrease with descent. Hope this helps!

5 0

2 years ago

Read 2 more answers

A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

Makovka662 [10]

Answer:

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k$

$a=5 i+2t j - 6\ t^2k$

Therefore the acceleration function a will be

$a=5 i+2t j - 6\ t^2k$

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

$a=\sqrt{5^2+4^2+24^2}\ m/s^2$

a= 24.83 m/s²

The direction of the acceleration a is given as

$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

5 0

2 years ago

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 0

2 years ago