The volume of an irregular object can be measured by

A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The str

alekssr [168]

Answer:

$F = 5253.7 N$

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

$T = \frac{mv^2}{L}$

now we have

$68 = \frac{m(16.53^2)}{3.7}$

now we have

$68 = 73.8 m$

$m = 0.92 kg$

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

$F = \frac{mv^2}{r}$

$F = \frac{0.92(71.5^2)}{0.896}$

$F = 5253.7 N$

8 0

3 years ago

Read 2 more answers

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = $\frac{1}{2} mV^{2} + mgh$

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In our situation

$\frac{0,71}{sin90} =\frac{h}{sin32}$ → $h=0,376$

Therefore

$1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376$

→ $V=2,416$

7 0

3 years ago

A bus travels east for 3 km the north for 4 km what is its final displacement

lesya [120]

Im about to the math for this right now.

8 0

2 years ago

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l

Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

Wₐ = 1500 lb

W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

W = mg

m = W / g

mₐ = Wₐ / g

m_b = W_b / g

mₐ = 1500/32 = 46.875 slug

m_b = 125/32 = 35,156 slug

Let's reduce to the english system

v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

p_f = (mₐ + m_b) v

the moment is preserved

p₀ = p_f

mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

v = $\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}$

we substitute the values

v = $\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6$

v = 0.559 v₀ₐ - 26.40 (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

M = mₐ + m_b

M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

K₀ = ½ M v²

final point. When they stop

K_f = 0

The work is

W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

N-W = 0

N = W

the friction force has the expression

fr = μ N

we substitute

-μ W x = Kf - Ko

-μ W x = 0 - ½ (W / g) v²

v² = 2 μ g x

v = $\sqrt{ 2 \ 0.750 \ 32 \ 17.5}$ Ra (2 0.750 32 17.5

v = 28.98 ft / s

3 0

2 years ago

Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

3 years ago