How much of a kilogram is a gram?

Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti

Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

7 0

2 years ago

ANYONE who is good with the consequences of Population growth PLEASE help me!!

alekssr [168]

in china, there is a family limit for only having 1 child

at 10 billion people on earth, we will most likely run out of food supply

4 0

2 years ago

Read 2 more answers

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

2 years ago

What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

In y-component:

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

In x-component:

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0

2 years ago

Yea, gonna need some help. Thanks

natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

$h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t$

where,

v₀ = initial speed = 110 ft/s

Therefore,

$110 = 32t\\\\t = \frac{110}{32}\\\\$

t = 3.48 s

8 0

2 years ago