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3 years ago

7

Which would be an example of biopsychology

1 answer:

Svetach [21]3 years ago

3 0

Answer:

It is just how your body works, like how your nervous system works. In other words, it is your structure and fuctions of cells in your body of the nervous system.

Hope this helps!

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The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhe

WARRIOR [948]

Answer:

6.8 mm

Explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet, $m_b=10 g=10\times 10^{-3} kg$

1 kg=1000 g

Total mass of system,M= $m+m_a=10+10\times 10^{-3}=10.01kg$

Speed of bullet,u=300 m/s

$\theta=30^{\circ}$

By law of conservation of momentum

$m_bucos\theta=Mv$

$v=\frac{m_bvcos\theta}{M}=\frac{0.01\times 300cos30^{\circ}}{10.01}=0.259m/s$

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

$\frac{1}{2}Mv^2-0=Mgh-0$

$\frac{1}{2}(10.01)(0.259)^2=10.01\times 9.8 h$

Where $g=9.8 m/s^2$

$h=\frac{(0.259)^2}{2\times 9.8}=0.0034m$

1m=100 cm

$h=0.0034\times 100=0.34 cm$

Distance traveled by block= $d=\frac{h}{sin\theta}=\frac{0.34}{sin30^{\circ}}=0.68 cm=6.8 mm$

1cm=10 mm

4 0

4 years ago

The electrical resistance of an element in a platinum resistance thermometer at 100°c, 0°c and room temperature are 75.00Ω, 63.0

Dominik [7]

Answer:

16.6 °C

Explanation:

From the question given above, the following data were obtained:

Temperature at upper fixed point (Tᵤ) = 100 °C

Resistance at upper fixed point (Rᵤ) = 75 Ω

Temperature at lower fixed point (Tₗ) = 0 °C

Resistance at lower fixed point (Rₗ) = 63.00Ω

Resistance at room temperature (R) = 64.992 Ω

Room temperature (T) =?

T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ

T – 0 / 100 – 0 = 64.992 – 63 / 75 – 63

T / 100 = 1.992 / 12

Cross multiply

T × 12 = 100 × 1.992

T × 12 = 199.2

Divide both side by 12

T = 199.2 / 12

T = 16.6 °C

Thus, the room temperature is 16.6 °C

6 0

3 years ago

When sound is created it travels to the ear through a

kaheart [24]

Sound travels through the ear via compression

6 0

3 years ago

Read 2 more answers

2. A 15 kg mass fastened to the end of a steel wire of un-

Flauer [41]

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

6 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago