PLEASE HELP ASAP!!!!!!!!!!

hypnotist covers a distance of 5 kilometre in one hour and Rahul covers the same distance in 2 hour calculate the speed and stat

adell [148]

Answer:

At 5km/hr; Hypnotist travels with a higher speed.

Explanation:

Let hypnotist be H while Raul be R.

Given the following data;

Distance H = 5km

Time H = 1 hour

Distance R = 5km

Time R = 2 hours

To find the their respective speed;

a. For H;

Speed = distance/time

Speed H = 5/1

Speed H = 5km/hr

b. For R;

Speed = distance/time

Speed R = 5/2

Speed R = 2.5 km/hr

Therefore, hypnotist travels with a higher speed.

3 0

2 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

3 0

3 years ago

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

Svetllana [295]

Answer:

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

Explanation:

We can find the gravitational potential energy using the following formula.

$GPE=mgh$

Identifying given information.

The nickel has a mass $m=0.005 \,kg$ , and it is a the top of Washington Monument.

The Washington Monument has a height of $h=555 \, ft$ , thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.