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4 years ago

8

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance.

2 answers:

stepladder [879]4 years ago

5 0

Answer:

t = 2.32 s

Explanation:

Applying the equation of motion;

d = ut + 0.5gt^2

Where;

d = distance travelled

u = initial velocity

g = acceleration due to gravity

t = time taken

Since the object was dropped;

u = 0

Then,

d = 0.5gt^2

t^2 = d/0.5g

t = √(d/0.5g) .......1

Given

g = 3.7 m/s^2

d = 10 m

Substituting the values;

t = √(10/(0.5×3.7))

t = 2.32 s

denpristay [2]4 years ago

4 0

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

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Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of

KonstantinChe [14]

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

$m_{s}c_{w} \alpha _{w}$ + $m_{s}l$ = $m_{w}c_{w} \alpha _{w}$ + $m_{c}c_{c} \alpha c_{c}$

L=specific latent heat of water(steam)=2268J/g

$c_{w}$ =specific heat capacity=4.2J/gK

$c_{c}$ =specific heat capacity of calorimeter =0.9J/gk

$m_{w}$ =280g

$m_{c}$ =38g

α=change in temperature

$\alpha _{c}$ =(40-25)=15

$\alpha _{w}$ =(40-25)=15

$\alpha _{s}$ =(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make $m_{s}$ the subject of formula

$m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}$

Hence

$m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g$

Hence the amount of steam needed is 7.2g

6 0

3 years ago

What is the weight of an astronaut who has a mass of 90 kg on the moon? (Note: acceleration due to gravity of the moon is 1.62 N

motikmotik

Answer:

W = 145.8 [N]

Explanation:

To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.

W = m*g

where:

m = mass = 90 [kg]

g = gravity acceleration = 1.62 [kg/m²]

W = 90*1.62

W = 145.8 [N]

8 0

3 years ago

slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

$v_f = 46.8 m/s$

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

$v_i = 15 m/s$

$g = - 9.81 m/s^2$

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

$4.905 t^2 - 15 t - 100 = 0$

so we have

$t = 6.3 s$

Now speed of the object when it reaches the ground is given as

$v_f = v_i + at$

$v_f = -15 + (9.81)(6.3)$

$v_f = 46.8 m/s$

8 0

4 years ago

A bullet is fired straight up from a gun with a

melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

$V=V_{o}+gt$ (1)

$V^{2}=V_{o}^{2}+2gy$ (2)

Where:

$V$ is the final velocity of the bullet

$V_{o}=152 m/s$ is the initial velocity of the bullet

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

$t=6.9 s$ is the time

$y$ is the vertical position of the bullet at $t=6.9 s$

Let's begin by finding $V$ from (1):

$V=152 m/s-9.8 m/s^{2}(6.9 s)$ (3)

$V=84.38 m/s$ (4)

Now we have to substitute (4) in (2):

$(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y$ (5)

Isolating $y$ :

$y=815.511 m$ This is the displacement of the bullet after 6.9 s

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3 years ago

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Elza [17]

N
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4 years ago

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