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3 years ago

The world’s largest gold bar, worth ten million dollars in 2014, has a base measuring 46 cm ×

Physics

1 answer:

LenaWriter [7]3 years ago

4 0

Answer:

a} 2400 N

b}240 kg

Explanation:

Pressure is force per unit area.

Given from the question that;

The pressure = 24000 Pa

Area = 0.46 * 0.22 =0.1012 m²

Calculating force from the formula;

P= F/A where P is pressure, F is force and A is area then

F= P*A = 24000*0.1012 = 2428.8 N

a) The weight is 2400 N in 2 significant figures

b) Mass is calculated from the formula;

w= m* g where w is weight, m is mass and g is acceleration due to gravity

2400 = m * 9.81

m= 2400/ 9.81

m= 244.64

m= 240 kg { in 2 significant figures}

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A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

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3 years ago

A piano tuner is tuning a piano when he discovers that the G above middle C is vibrating with a higher frequency than his G tuni

kotegsom [21]

Answer:

Tuning a piano is the process of adjusting the tension of its strings, thereby altering their pitch, or frequency of vibration, by slightly turning the tuning pins to which they're attached, so that each string sounds pleasingly in harmony with every other string.

<h2>Hopefully u will satisfy with my answer of ur question..!!</h2>

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3 years ago

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe

Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet = 327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

= 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

= 0.00353 × 128 × (327-40)

= 0.00353 × 128 × 287

= 129.68 J

Quantity of heat required to melt the bullet:

$Q = mH_f$

Q = 0.00353 × 24500

= 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

= 216.17 J

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3 years ago

A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?

AlexFokin [52]

The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.

At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly

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