Calculate the mass of a liter of water in grams.

A hammer of mass M is moving at speed v0 when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer i

OleMash [197]

Answer:

i think it would be D

8 0

2 years ago

A high-performance sports car can go from 0 to 100 mph in 5.8 s. (Assume the car travels in the positive direction. Indicate the

sasho [114]

Answer: a) 7.71 m/s², b) - 6.67 m/s²

Explanation: first thing to note is that

1 mile = 1609.34

1 hour = 3600s

Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s

Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?

By using newton's laws of motion

v = u + at

44.71 = 0 + a(5.8)

44.71 = 5.8a

a = 44.71/5.8

a = 7.71 m/s²

Question b)

The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.

Hence u = 34 m/s, v = 0, a =?, t = 5.1 s

v = u + at

0 = 34 + a(5.1)

a(5.1) = - 34

a = - 34/5.1

a = - 6.67 m/s².

The negative sign beside the acceleration shows that the body is decelerating

7 0

3 years ago

I need help on this still stuck on it

jenyasd209 [6]

The underlined offset 2 is called a subscript and it describes how many of the element preceding it per molecule of the entire molecule/compound. In the case of H20, it means there are 2 Hydrogen per H20 molecule.

6 0

2 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s

ziro4ka [17]

A) Initial velocity of ball 1: 9.53 m/s upward

B) Height of the building: 6.48 m

C) Maximum velocity: 11.7 m/s

D) Minimum velocity: 5.83 m/s

Explanation:

A)

The y-position of the 1st ball at time t is given by the equation for free fall motion:

$y_1 = h + v_0 t - \frac{1}{2}gt^2$ (1)

where

h = 21.0 m is the initial height of the ball, the height of the building

$v_0$ is the initial velocity of the ball, upward

$g=9.8 m/s^2$ is the acceleration of gravity

The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by

$y_2 = h-\frac{1}{2}g(t-1.19)^2$

where

h = 21.0 m is the initial height of the ball, the height of the building

t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then $y_2=h$ , so the ball is still on the roof

The 2nd ball reaches the ground when $y_2=0$ , so:

$0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0$

Which has two solutions:

t = -0.88 s (negative, we discard it)

t = 3.26 s (this is our solution)

The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and $y_1=0$ , so we find the initial velocity:

$0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s$

B)

In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:

$v_0 = 8.70 m/s$

When the two balls reach the ground at the same time, there position is the same, so we can write:

$y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2$

Solving the equation, we find:

$v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s$

This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:

$0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m$

C)

If $v_0$ is greater than some value $v_{max}$ , then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}$

becomes negative: in that case, the time becomes negative, so no solution is possible.

The denominator becomes negative when

$1.19g-v_0 < 0$

Therefore when

$v_0>1.19g=(1.19)(9.8)=11.7 m/s$

So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.

D)

There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19$