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3 years ago

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Calculate the magnitude of the force, in newtons, exerted by a boxing glove on an opponent’s face, if the glove and face togethe

r compress 7.35 cm during a blow in which the 7.25-kg arm and glove are brought to rest from an initial speed of 10.5 m/s.

1 answer:

saul85 [17]3 years ago

7 0

Answer:

5437.5 N

Explanation:

from the question we are given the following:

mass = 7.25 kg

initial speed (u) = 10.5 $\frac{m}{s}$

finial velocity (v) = 0

compression (s) = -7.35 cm = -0.0735 m (the negative sign is because the face is compressed, so there is a decrease)

we know that force = mass x acceleration,

we have our mass but need to find the acceleration. we can get the acceleration by applying the formula below

$v^{2} = u^{2} + 2as$

therefore

0 = 10.5^{2} + (-2)× a × 0.0735[/tex]

a = $\frac{10.5^{2} }{ 2 × 0.0735$ }[/tex]

a = 750

from F = m x a

force = 7.25 x 750 = 5437.5 N

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In a parallel circuit the current A stays in one path B splits and goes through two components C makes one circle

ArbitrLikvidat [17]

Answer:

B splits and goes through two components

Explanation:

- A series circuit is a circuit in which the components are all connected along the same branch: as a result, the current flowing through the components is the same, while the sum of the potential differences across each component is equal to the emf of the battery

- A parallel circuit is a circuit consisting of separate branches, so that each branch has a potential difference equal to the emf of the battery. As a result, in such a circuit the current in the circuit splits and goes through the different branches/components.

So, the correct answer is

B splits and goes through two components

8 0

3 years ago

Which characeristic makes omasis different from diffusion.

Ksju [112]

<span>haha I used to think biology was so hard, i find it quite easy now.
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OSMOSIS, the important thing to remember is that water ALWAYS flow towards the region with the higher concentration of the solute (ex: Salt is solute, water is solvent) solute is the thing that is being dissolved. Solvent is the one doing the dissolving. Hope this helped!</span>

5 0

3 years ago

Please help on this one?

bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i

pantera1 [17]

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

<u>Δx = 4.68 x 10⁻³ m = 4.68 mm</u>

5 0

3 years ago