A car covers a distance of 200m. If its velocity is 20 m/s, calculate the time taken.

A piece of a metal alloy with a mass of 114 g was placed into a graduated cylinder that contained 25.0 mL of water, raising the

Minchanka [31]

Answer:

6.51 g/c.c

Explanation:

mass, m = 114 g

initial volume, V1 = 25 mL

final volume, V2 = 42.5 mL

Volume of the metal piece, V = V2 - V1 = 42.5 - 25 = 17.5 mL

1 mL = 1 c.c

So, Volume of metal, V = 17.5 c.c.

Let the density of the metal is d.

density = mass / volume

d = 114 / 17.5 = 6.51 g/c.c

Thus, the density of metal is 6.51 g/c.c.

8 0

3 years ago

what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this

natali 33 [55]

Answer:

55 miles / hour = 1.475232 kilometers / minute

6 0

3 years ago

Read 2 more answers

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

4 years ago

A heavy steel cable attached to a motor is lifting a girder. The girder is speeding up.

Yanka [14]

Answer: part a) both the tension and gravity vectors will be pointing downwards.

part b) the tension vector will be pointing upwards and the gravity will be pointing downwards.

Explanation:

part a) Since you are looking at the FBD for the steel cable here, the tension will be pointing downwards because the girder is hung from the cable which creates a downwards tension. Gravity will always point downwards, to the ground.

part b) Since you are looking at the FBD for the girder here, the tension will be pulled upwards because the cable is weighted which creates an upwards tension. Gravity will always point downwards, to the ground.

Sorry for the poor explanation, but hopefully this makes sense.

8 0

3 years ago

An astronaut in a satellite 1600km above the Earth experience a gravitational force magnitude 700N on earth.The Earth's radius i

Brilliant_brown [7]

Answer:

The magnitude of the gravitational force which the astronaut experience in satellite $= 448$ Newton