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mr Goodwill [35]
3 years ago
12

a man checked himself into seattle hospital. he didnt know who he was and had no idea how he got to seattle. the hospital staff

could find nothing wrong with him and sent his picture to tv stations. a relative who lived in england recognized the man, and came to get him. the man was probably suffering from
Physics
2 answers:
8090 [49]3 years ago
7 0

Amnesia would be the answer

marin [14]3 years ago
7 0

Answer: Amnesia.

Explanation: Here we see the case of a man that has no short lapse memory nor long lapse memory, you can infer this because the man does not know his name and does not know why he is where he is.

This can be a case of amnesia, which is a deficiency in the memory caused by brain damage or disease. This can be temporal or total, and can only affect selected sections of the memory.

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The time it takes two successive crests of an ocean wave to pass a given point is called a _____.
nika2105 [10]
The answer is Period
3 0
3 years ago
Read 2 more answers
Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

6 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
A child throws a baseball upward with an initial velocity of 20 m/s. The child wants to throw the baseball at least as high as t
Umnica [9.8K]
When the ball starts its motion from the ground, its potential energy is zero, so all its mechanical energy is kinetic energy of the motion:
E= \frac{1}{2}mv^2
where m is the ball's mass and v its initial velocity, 20 m/s.

When the ball reaches its maximum height, h, its velocity is zero, so its mechanical energy is just gravitational potential energy:
E=mgh

for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
\frac{1}{2}mv^2 = mgh
From which we find the maximum height of the ball:
h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m

Therefore, the answer is yes, the ball will reach the top of the tree.

5 0
3 years ago
A force of 25N acts on a mass of 5.0kg, initially at rest. Calculate the distance travelled before achieving a velocity of 20m/s
spin [16.1K]

Answer:

40m

Explanation:

let's calculate the acceleration first

force = mass × acceleration

rearranging to find acceleration:

acceleration = force ÷ mass

force = 25N, mass = 5.0kg

acceleration = 25 ÷ 5 = 5ms^-2

we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance

rearranging v^2 = u^2 + 2as the distance is

s = (v^2 - u^2) ÷ 2a

v = 20, u = 0, a = 5

s = (20^2 - 0^2) ÷ (2 × 5) = 40m

the distance is 40m

6 0
2 years ago
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