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3 years ago

17. A skateboard has a mass of 3.2 kg. If it slips out from a skateboarder and has

Physics

1 answer:

Akimi4 [234]3 years ago

6 0

<h2>Book will hit the foot with force 21.8 N</h2>

Explanation:

The momentum of any object = m v

here m is the mass of body and v is its velocity

Thus momentum P = 3.2 x 1 = 3.2 kg-m

As the book is falling freely , so it will fall with and acceleration g

The force exerted by it is = m g

Thus force = 2.18 x 10 = 21.8 N

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Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

Romashka [77]

Answer:

E = 5291.00 N/C

Explanation:

Expression for capacitance is

$C = \frac{\epsilon A}{d}$

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

$C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}$

$C = 24.06 \epsilon$

$C = 24.06\times 8.854\times 10^{-12} F$

$C =2.1\times 10^{-10} F$

We know that capacitrnce and charge is related as

$V = \frac{Q}{C}$

$= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}$

v = 9.523 V

Electric field is given as

$E = \frac{V}{d}$

= $\frac{9.52}{1.8*10^{-3}}$

E = 5291.00 V/m

E = 5291.00 N/C

5 0

3 years ago

People in finance work mostly with: <br> Insurance <br> Products <br> Information <br> Money

zloy xaker [14]

The answer is money. Hope this helps.

6 0

3 years ago

Read 2 more answers

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

7 0

3 years ago

Find the image position for a picture placed 3.0 cm outside the focal point of a converging lens with a 4.0 cm focal length. a.

horrorfan [7]

<span>Answer: Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance) 1/4 = 1/7 + 1/d 1/4 - 1/7 = 1/d 3/28 = 1/d d = 28/3 d = 9.33 cm</span>

5 0

4 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$

The volume of water is:

$139g.\frac{1mL}{0.997g} =139mL$

7 0

4 years ago