Stress is a factor that contributes to heart disease risk.true or false

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a) Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 1.5 m

Substituting

s = ut + 0.5 at²

1.5 = 0 x t + 0.5 x 9.81 xt²

t = 0.553 s

Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 200 m/s

Acceleration, a = 0 m/s²

Time, t = 0.553 s

Substituting

s = ut + 0.5 at²

s = 200 x 0.553 + 0.5 x 0 x 0.553²

s = 110.6 m

The bullet travels horizontally 110.6 m

6 0

2 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

SVETLANKA909090 [29]

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$

5 0

3 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

3 years ago

Activity

vodka [1.7K]

Your welcome LOL plz like

5 0

2 years ago

The brakes on a 15,680 N car exert a stopping force of 640 N.

Murrr4er [49]

Answer:

1568 Kg is the answer

Explanation:

6 0

3 years ago