All categories

2 years ago

3. The wind exerts a force of 452 N north on a sailboat, while the water

Physics

1 answer:

maksim [4K]2 years ago

7 0

Explanation:

The two forces form a right triangle.

The magnitude is found with Pythagorean theorem.

F² = Fx² + Fy²

F² = (-325 N)² + (452 N)²

F = 557 N

The direction is found with trigonometry:

θ = atan(Fy / Fx)

θ = atan(452 N / -325 N)

θ = -54.3° or 125.7°

θ is 125.7° from the +x axis (east). That corresponds to a direction of 35.7° west of north, or 54.3° north of west.

The net force has a magnitude of 557 N and a direction of 35.7° west of north.

You might be interested in

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

= surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ / 2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀

= 6.78 X 10³ N/C

4 0

2 years ago

1. Which of the following regarding a collision is/are true? a. If you triple your speed your force of impact will be three time

34kurt

Answer:

Explanation:

If you double your speed, the energy dissipated in a crash is four times greater

Because impact increases with square of increase in speed.

7 0

2 years ago

Read 2 more answers

A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,

Alik [6]

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

<u>Explanation</u>:

By the definition of the coefficient of static friction we have:

μ $_{s}$ = $\frac{F_{appl} }{W}= \frac{F_{s} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{s}$ is the static friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to just start the crate moving:

$F_{appl} = F_{s} = u_{s}N = u_{s}mg$

= 0.760 $\times$ 60 kg $\times$ 9.8 m / s^2

= 447 N.

By the definition of the coefficient of kinetic friction we have:

u $_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{k}$ is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

$F_{appl} = F_{k} = u_{k}N = u_{k}mg$

= 0.410 $\times$ 60 $\times$ 9.8

= 241 N.

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

6 0

2 years ago

In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated. Wha

k0ka [10]

Answer:

Temperature of the sink will be 191.1 K

Explanation:

We have given that heat withdrawn form the source = 10 KJ

Work done = 3 KJ

We know that efficiency is given by

$\eta =\frac{work\ done}{heat\ withdrawn}=\frac{3}{10}=0.3$

Higher temperature is given by $T_1=273K$

We have to find the lower temperature $T_2$

We know that efficiency is also given by

$1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$

So $\frac{273-T_2}{273}=0.3$

$T_2=191.1K$

So temperature of the sink will be 191.1 K

5 0

2 years ago

Read 2 more answers