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zloy xaker [14]

2 years ago

14

When a nuclear power accident occurs: all of the effects are seen immediately. the effects are powerful yet fleeting. only a sma

ll radius surrounding the accident is affected. radiation may stick around for a long time and cause cancer.

2 answers:

alekssr [168]2 years ago

8 0

Out of the four the last one is true - "<span>radiation may stick around for a long time and cause cancer". The other three are false - some effects will be long-lasting, they may affect large areas, and are not necessarily apparent immediately.</span>

rewona [7]2 years ago

3 0

Answer:

radiation may stick around for a long time and cause cancer

Explanation:

radiation may stick around for a long time and cause cancer

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What was Great Britain’s policy using African Americans during the war

kifflom [539]

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Explanation:

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2 years ago

510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr

Yuki888 [10]

Answer:

The terminal velocity is $v_t =17.5 \ m/s$

Explanation:

From the question we are told that

The mass of the squirrel is $m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg$

The surface area is $A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2$

The height of fall is h =4.8 m

The length of the prism is $l = 23.2 = 0.232 \ m$

The width of the prism is $w = 11.6 = 0.116 \ m$

The terminal velocity is mathematically represented as

$v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }$

Where $\rho$ is the density of a rectangular prism with a constant values of $\rho = 1.21 \ kg/m^3$

$C$ is the drag coefficient for a horizontal skydiver with a value = 1

A is the area of the prism the squirrel is assumed to be which is mathematically represented as

$A = 0.116 * 0.232$

$A = 0.026912 \ m^2$

substituting values

$v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }$

$v_t =17.5 \ m/s$

7 0

3 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

2 years ago