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Arte-miy333 [17]

2 years ago

13

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

force of 16 N. What is the net force on them?

2 answers:

sammy [17]2 years ago

6 0

Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, $F_1 = 32\ N$ (right side)

Force acting by Simone , $F_2 = -16\ N$ (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

So, the net force is 16 N and it is acting to the right side.

belka [17]2 years ago

5 0

Answer:

16 N (towards right)

Explanation:

Pulling force towards right, F1 = 32 N

Pulling force towards left, F2 = 16 N

Net force = F1 - F2

F = 32 - 16 = 16 N (towards right)

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

2 years ago

Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

$\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}$

here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0

2 years ago

Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x

never [62]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

8 0

2 years ago

Read 2 more answers

The wavelength of green light is 550 nm.

SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0

2 years ago

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago