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3241004551 [841]

3 years ago

5

explain why when someone uses his thumb to push a pin into a block of wood the pressure on the wood is greater than the pressure

on this thumb

2 answers:

Charra [1.4K]3 years ago

8 0

Answer:

the smaller the area the bigger the pressure

NeTakaya3 years ago

7 0

Because the tip of the pin has a very small area, therefore greater pressure. the area of the pin pressing on the thumb is much greater.

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ASAP answer all pls will mark brainiest

sammy [17]

Answer:

1) B

2)D

Explanation:

I am unsure of #2

7 0

2 years ago

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As a 3.0 kg bucket is being lowered into a 10 m deepwell, starting from top, the tension in the rope is 9.8 N. theacceleration o

777dan777 [17]

Answer:

A) 6.5 m/s²

Explanation:

Mass of the bucket, m = 3.0 kg

depth of the well, d = 10 m

tension on the rope, T = 9.8 N

The net downward force on the bucket is given as;

T = mg - ma

where;

a is downward acceleration of the bucket

9.8 = (3 x 9.8) - 3a

9.8 = 29.4 - 3a

3a = 29.4 - 9.8

3a = 19.6

a = 19.6 / 3

a = 6.53 m/s² downwards

Therefore, the acceleration of the bucket is 6.53 m/s² downwards

8 0

3 years ago

A 15 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 53

blagie [28]

Answer:

0.0613°C

Explanation:

the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s

the bullet moves with different kinetic energies before and after the penetration, therefore

Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)

= $\frac{1}{2}$ × 15×10⁻³ × (865² - 534²)

= 3.47 × 10⁻³J

this loss in energy is transferred to the water, therefore

change in temperature = $\frac{Q}{m C}$

where c = heat capacity of water = 4.19 x 10^3

m = mass of water = 13.5 kg

= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }

=0.0613°C

5 0

3 years ago

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After t

Pavlova-9 [17]

Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
$a = - 2.5 {ms}^{ - 2}$

$t = 4s$
$v = 3.0 {ms}^{ - 1}$
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$v = u + at$
we can determine the initial velocity

$3.0= u + -2.5(4)$

$3.0+2.5(4)=u$

$13.0 = u$

Hence the initial velocity is 13.0 meters per seconds

3 0

3 years ago

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago