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Bingel [31]
2 years ago
13

Crystal can ride her bike 10 miles in 2.1 hours. What is her velocity

Physics
1 answer:
masha68 [24]2 years ago
6 0

Answer:

v = 4.76 m/s

Explanation:

Given,

The distance traveled by her bike, d = 10 miles

The time of her travel, t = 2.1 m/s

The velocity of an object is defined as the distance traveled by the object to the time of travel. Therefore,

                          V = d/t   m/s

                              = 10 / 2.1

                              = 4.76 m/s

Hence, The velocity of her bike is, V = 4.76 m/s

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0
3 years ago
Four importance of soil water
Andru [333]

Answer:

nitrogen, phosphorus, potassium, and calcium

Explanation:

Just put it.

4 0
2 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an
jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0
3 years ago
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
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