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Nimfa-mama [501]

2 years ago

10

a radioactive tracer has a half-life of 2 hours. how much of a 2500g sample will be avalible after 18 hours

1 answer:

zhannawk [14.2K]2 years ago

8 0

Answer:

i believe the answer is 2.44140625 grams remaining

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If the photon scatters in the backward direction, what is the magnitude of the linear momentum of the electron just after the co

Strike441 [17]

Momentum is a vector quantity, and is always conserved. Whenever a collision occurs between two objects, the objects behave under the principle of conservation of momentum. Therefore, if an object moves in the direction opposite to its original direction after a collision, then this indicates that the momentum of the colliding object was greater than the object under consideration.

8 0

2 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

7 0

3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

A wave is traveling from one medium to another.If the wave is partially transmitted an partially reflected from the boundary,wha

Kamila [148]

Answer: the answer is C

Explanation:

3 0

2 years ago

ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,

Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

$d = \frac{(v_f + v_i)}{2} \times t$

here we know that

$v_f = 10 m/s$

$v_i = 0$

$t = 1.2 s$

now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

4 0

2 years ago