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sergiy2304 [10]
3 years ago
9

A ball is thrown vertically downward at 10 m/s. what is it’s speed 1s and 2s later

Physics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

20 m/s

30 m/s

Explanation:

Given:

v₀ = -10 m/s

a = -9.8 m/s²

When t = 1 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (1 s)

v = -19.8 m/s

When t = 2 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (2 s)

v = -29.6 m/s

Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.

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Which nucleus completes the following equation?
IgorLugansk [536]

Answer: ^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr

The chromium nucleus will complete the reaction.

Explanation:

The given reaction is a type of radioactive nuclei decay:

Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as \beta ^+-decay. In this the mass number remains same.

_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

Now according to the given reaction:

^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr

The chromium nucleus will complete the reaction.

4 0
3 years ago
Read 2 more answers
A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
IrinaK [193]

Answer : The specific heat capacity of the alloy 1.422J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of alloy = ?

C_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of alloy = 21.6 g

m_2 = mass of water = 50.0 g

T_f = final temperature of system = 31.10^oC

T_1 = initial temperature of alloy = 93.00^oC

T_2 = initial temperature of water = 22.00^oC

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

21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC

c_1=1.422J/g^oC

Therefore, the specific heat capacity of the alloy 1.422J/g^oC

6 0
3 years ago
What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?
sveta [45]
Momentum = (mv). 
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
8 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
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