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3 years ago

• 2. particles penetrate through paper.

1 answer:

7 0

Answer: The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper and Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.

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Could I get help on this question please . My parents won’t help me /:

vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

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

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

3 0

3 years ago

Wind erosion can be reduced by _____.

yanalaym [24]

Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force

4 0

3 years ago

If the ankylosaurs starts running and accelerates at 1.3m/s^2 for 3 seconds how far does he make it before the velociraptor catc

sveta [45]

Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.

Acceleration = 1.3 m/s²

Velocity: ∫ 1.3 dx = 1.3x + c m/s

Distance: ∫ 1.3x dx = 1.3x²/2 + c m

Distance run: 1.3*3²/2 = 5.85 m

What bad thing happened?

7 0

3 years ago

While John is traveling along a straight interstate

lora16 [44]

Answi dont know

Explanation:

i dont know

7 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

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

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago