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3 years ago

How is impact force and impact time related when describing the same impulse?

Physics

1 answer:

Radda [10]3 years ago

7 0

Answer:

The correct option is (a)

Explanation:

Impulse is given by the product of force and time. It is denoted by J. So,

$J=F\times t$

$F=\dfrac{J}{t}$

It can be seen that the relation between the force and the impact time is inverse. Hence, the impact force and impact time possess inverse relation.

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If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit

Oxana [17]

Answer:

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

4 0

3 years ago

What new characteristics did john dalton add to the model of the atom

kicyunya [14]

Answer:

All atoms of an element are identical. The atoms of different elements vary in size and mass. Compounds are produced through different whole-number combinations of atoms. A chemical reaction results in the rearrangement of atoms in the reactant and product compounds.

HOPE THIS HELPED!!!!!!!!!!!!!!!!!!! XDDDDDDDDDDD

7 0

3 years ago

A girl is out jogging at 2.00 m/s and accelerates at 1.50 m/s^2 until she reaches a velocity of 5.00 m/s. How far does she get?

Maksim231197 [3]

Answer:

7.00 m

Explanation:

Given:

v₀ = 2.00 m/s

v = 5.00 m/s

a = 1.50 m/s²

Find: Δx

v² = v₀² + 2aΔx

(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx

Δx = 7.00 m

5 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

As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?

Y_Kistochka [10]

Answer:

C. Angle of Attack.

Explanation:

The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.

And thus, we have

Lift = Weight

3 0

3 years ago