All categories

3 years ago

Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?

Physics

1 answer:

Mkey [24]3 years ago

4 0

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F = (1200)(0) = 0

You might be interested in

Which of the following statements is always true?

Gwar [14]

Answer: B

Explanation:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

7 0

3 years ago

You pull two blocks connected by a lightweight string across a horizontal table. Block 1 has a mass of 1.00 kg, and block 2 has

coldgirl [10]

Use google
-Just sayi’n don’t get mad

7 0

3 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

The work accomplished (produced) by a machine is called Work _____

Nonamiya [84]

I think it’s output because output work is work done by a machine

6 0

3 years ago

If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.

Schach [20]

Answer:

The magnitude of each charge is $2.82\times10^{-6}\ C$

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Where, q = charge

r = separation

Put the value into the formula

$20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}$

$q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}$

$q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}$

$q=2.82\times10^{-6}\ C$

Hence, The magnitude of each charge is $2.82\times10^{-6}\ C$

6 0

3 years ago

Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​

Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?