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3 years ago

1. A brick and a peanut are dropped of the roof of a house at the same time, which

Physics

1 answer:

iris [78.8K]3 years ago

3 0

Answer:

It would be the brick since the mass and the weight is greater than the peanut I might be wrong on this but thats the best answer i can give

Explanation:

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In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

3 years ago

A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i

Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft and d(5) = 35 ft

$V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s$

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft and d(2.1) = 8.61 ft

$V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s$

4 0

3 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

Read 2 more answers

Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28

Ludmilka [50]

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{28.1\text{ years}}$

$k=2.47\times 10^{-2}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $2.47\times 10^{-2}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant = 1.00 g

a - x = amount left after decay process = 0.600 g

Now put all the given values in above equation, we get

$t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}$

$t=20.7\text{ years}$

Therefore, the time passed in years is 20.7 years.

8 0

3 years ago

Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain

anyanavicka [17]

I think these two variables are sufficient to completely specify the state.

In an isolated room with air only ,the volume is fixed.Mass ,density and its specific volume can be easily known.

Other thermodynamic properties like entropy, enthalpy etc are also fixed at a given temperature & pressure.

4 0

3 years ago