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ElenaW [278]
4 years ago
12

A sled with no initial velocity accelerates at rate of 5.0 m/s^2 down a hill . How long does it take the sled to go 45 m to the

bottom
Physics
1 answer:
Leya [2.2K]4 years ago
6 0

The time taken by the sled to reach 45 m to the bottom is 4.24 s.

The sled's initial velocity <em>u</em> is zero. It has an acceleration <em>a</em> of a value 5 m/s² down the hill and it travels a distance <em>s</em> equal to 45 m down the hill.

Use the equation of motion,

s=ut+\frac{1}{2} at^2

Substitute 0 m/s for u and rewrite the equation for t.

t=\sqrt{\frac{2s}{a}}  \\ =\sqrt{\frac{2(45 m)}{5.0m/s^2} } \\ =4.24s

The sled takes a time of 4.24 s to reach the bottom of the hill

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8 0
1 year ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
3 years ago
Explain how tidal forces are causing Earth to slow down.
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6 0
3 years ago
When two objects with electrical charges interact, which affect the strength of that interaction?
noname [10]
The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span> 
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3 years ago
A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app
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Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

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E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

E = 995.12 N/C

8 0
3 years ago
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