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3 years ago

The space shuttle is descending through the earth's atmosphere. How is the force of gravity affected?

Physics

2 answers:

polet [3.4K]3 years ago

5 0

The answer would be C. It will decrease with descent. Hope this helps!

Lady_Fox [76]3 years ago

4 0

The answer should be It will decrease with descent

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A car moves with a speed of 72 km/h for 15 minutes and then with a speed of 80 km/h for the next 12 minutes. The total distance

sukhopar [10]

Answer:

The total distance covered by the car is 3,810.08 m

Explanation:

Given;

initial speed of the car, u = 72 km/hr = 20 m/s

initial time, t₁ = 15 minutes = 900 s

final speed of the car, v = 80 km/hr = 22.22 m/s

final time, t₂ = 12 minutes = 720 s

The acceleration of the car is given as;

$a = \frac{v-u}{t_2 -t_1} \\\\a = \frac{22.22-20}{900-720}\\\\a = 0.0123 \ m/s^2$

The total distance covered by the car is given as;

v² = u² + 2as

where;

s is the total distance covered by the car

22.22² = 20² + 2(0.0123)s

22.22² - 20² = 2(0.0123)s

93.728 = 0.0246s

s = 93.728 / 0.0246

s = 3,810.08 m

Therefore, the total distance covered by the car is 3,810.08 m

8 0

3 years ago

Read 2 more answers

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s

8 0

3 years ago

An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?

never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

<span>Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.</span>

6 0

3 years ago

If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.

Alex Ar [27]

The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

Hope it helped!

5 0

3 years ago

g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e

rusak2 [61]

Answer:

$(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0$

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

$(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0$

Given the function:

$p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}$

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in $[0,\infty).$

(2)

$\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1$

The function p(x) satisfies the conditions for a probability density function.

6 0

3 years ago