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3 years ago

How far does a sled travel in 5 seconds while accelerating from 4 m/s to 10 m/s?

Physics

2 answers:

Basile [38]3 years ago

8 0

Answer:

35 m

Explanation:

Given:

v₀ = 4 m/s

v = 10 m/s

t = 5 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (10 m/s + 4 m/s) (5 s)

Δx = 35 m

mina [271]3 years ago

3 0

Answer:

All we need to know for this is SUVAT equations. There are 5 SUVAT equations you need to learn. S = distance, U = initial velocity, V = final velocity, A = acceleration, T = time V= U + AT V^2 = U^2 + 2AS S = UT + 1/2AT^2 S = VT -

Explanation:

don't now if this hellps or not

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Irina18 [472]

Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

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5 0

3 years ago

A tall cylinder with a cross-sectional area 12.0 cm² is partially filled with mercury; the surface of the mercury is 5.00 cm abo

JulijaS [17]

Answer:

$V = 816 cm^3$

Explanation:

As we know that gauge pressure of the fluid at the bottom of the cylinder is given as

$P = \rho g h$

now we know that pressure at the bottom is double when water is poured on the mercury

So we have

$\rho_{hg} g h_1 = \rho_w g h_2$

so we will have

$13.6 \times 5 = 1 \times h$

so we have

$h = 68 cm$

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4 years ago

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of eac

Naddik [55]

Answer:

t = (ti)ln(Ai/At)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Explanation:

Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t

At = Ai/2^n ..... 1

Where n is the number of half-life that have passed.

n = t/half-life

Half life = 14

n = t/14

At = Ai/2^(t/14)

From equation 1.

2^n = Ai/At

Taking the natural logarithm of both sides;

nln(2) = ln(Ai/At)

n = ln(Ai/At)/ln(2)

Since n = t/14

t/14 = ln(Ai/At)/ln(2)

t = 14ln(Ai/At)/ln(2)

Ai = 800

At = 50

t = 14ln(800/50)/ln(2)

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Solving for t

t = 14×4 = 56 seconds

Let half life = ti

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4 years ago

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